[guided]The difficulty is that a determinant is a product of weakly convergent gradients, and products of weakly convergent sequences do not usually converge. The null-Lagrangian identity avoids this problem by moving one derivative onto the fixed test function and leaving one factor of $u_k$ undifferentiated, where compactness gives strong convergence.
We prove the assertion by induction on the size $s$ of the minor. When $s=1$, the minor has the form $\partial_{x_{b_1}}(u_k)_{a_1}$. Since weak convergence in $W^{1,p}(\Omega;\mathbb R^m)$ includes weak convergence of every first weak derivative in $L^p(\Omega)$, we have
\begin{align*}
\partial_{x_{b_1}}(u_k)_{a_1}\rightharpoonup \partial_{x_{b_1}}u_{a_1}
\end{align*}
in $L^p(\Omega)$, hence also in distributions.
Now assume the result has been proved for all minors of size $s-1$, with $2\le s\le r$. Fix a row tuple $A=(a_1,\dots,a_s)$, a column tuple $B=(b_1,\dots,b_s)$, and a test function $\varphi\in C_c^\infty(\Omega)$. Choose a bounded Lipschitz open set $V$ such that $\operatorname{supp}\varphi\subset V$ and $\overline V\subset\Omega$. This localization keeps all integrations away from the boundary of $\Omega$, so [integration by parts](/theorems/210) produces no boundary term.
For each $1\le b\le s$, define the lower-order minor
\begin{align*}
C_{b,k}=M_{(a_2,\dots,a_s),B_b}(Ju_k),
\end{align*}
where $B_b$ is the column tuple obtained by deleting $b_b$. Define $C_b$ in the same way with $u$ in place of $u_k$. The induction hypothesis applies to $C_{b,k}$ because it is an $(s-1)$-minor. Since $p\ge s$, we have $p>s-1$, so the stronger part of the induction hypothesis gives
\begin{align*}
C_{b,k}\rightharpoonup C_b \quad \text{in } L^{\rho_{s-1}}(V), \qquad \rho_{s-1}=\frac{p}{s-1}.
\end{align*}
We also need a strongly convergent factor to pair with this weakly convergent lower-order minor. The factor supplied by the divergence identity is $(u_k)_{a_1}\partial_{x_{b_b}}\varphi$. By local compactness, [citetheorem:8731] gives strong convergence of $u_k$ to $u$ in the required local Lebesgue space after we check the exponent. The correct exponent is
\begin{align*}
q_s=\frac{p}{p-s+1},
\end{align*}
because it is conjugate to the exponent $\rho_{s-1}$. Define
\begin{align*}
\rho_{s-1}=\frac{p}{s-1}.
\end{align*}
Then
\begin{align*}
\frac{1}{q_s}+\frac{s-1}{p}=1.
\end{align*}
Since $p\ge s$, one has $1\le q_s\le p$. If $p<n$, then $q_s\le p<p^*$; if $p=n$, then $q_s$ is finite; if $p>n$, the Morrey case of [citetheorem:8731] gives compact convergence into continuous representatives and hence into $L^{q_s}(V)$. Compactness first gives strong convergence for subsequences, but the weak limit in $W^{1,p}(V;\mathbb R^m)$ is fixed as $u|_V$, so every compactly convergent subsequence has strong limit $u|_V$. The subsequence contradiction criterion therefore yields
\begin{align*}
u_k|_V\to u|_V \quad \text{in } L^{q_s}(V;\mathbb R^m).
\end{align*}
Since $\partial_{x_{b_b}}\varphi$ is smooth and compactly supported in $V$, multiplication by $\partial_{x_{b_b}}\varphi$ preserves strong convergence in $L^{q_s}(V)$. Hence
\begin{align*}
(u_k)_{a_1}\partial_{x_{b_b}}\varphi\to u_{a_1}\partial_{x_{b_b}}\varphi
\end{align*}
strongly in $L^{q_s}(V)$.
We now use the elementary weak-strong product principle: if $f_k\to f$ strongly in $L^{q_s}(V)$ and $g_k\rightharpoonup g$ weakly in $L^{p/(s-1)}(V)$, then
\begin{align*}
\int_V f_k(x)g_k(x)\,d\mathcal L^n(x)\to\int_V f(x)g(x)\,d\mathcal L^n(x).
\end{align*}
Indeed, write the difference as
\begin{align*}
\int_V (f_k-f)(x)g_k(x)\,d\mathcal L^n(x)+\int_V f(x)(g_k-g)(x)\,d\mathcal L^n(x).
\end{align*}
The first term tends to $0$ by Holder's inequality and boundedness of $(g_k)$ in $L^{p/(s-1)}(V)$; the second tends to $0$ by weak convergence of $g_k$ against the fixed function $f\in L^{q_s}(V)$.
Applying this with
\begin{align*}
f_k=(u_k)_{a_1}\partial_{x_{b_b}}\varphi,\qquad g_k=C_{b,k},
\end{align*}
gives convergence of every term in the divergence identity:
\begin{align*}
\lim_{k\to\infty}\int_V (u_k)_{a_1}(x)\partial_{x_{b_b}}\varphi(x)C_{b,k}(x)\,d\mathcal L^n(x)=\int_V u_{a_1}(x)\partial_{x_{b_b}}\varphi(x)C_b(x)\,d\mathcal L^n(x).
\end{align*}
We now record the divergence identity used for the tested $s$-minor. For each Sobolev map $v\in W^{1,p}(V;\mathbb R^m)$ and each $1\le b\le s$, define
\begin{align*}
D_b(v)=M_{(a_2,\dots,a_s),B_b}(Jv).
\end{align*}
For smooth $v$, expanding the determinant along the first row gives
\begin{align*}
M_{A,B}(Jv)=\sum_{b=1}^s(-1)^{1+b}\partial_{x_{b_b}}v_{a_1}D_b(v).
\end{align*}
The cofactor fields satisfy the Piola cancellation
\begin{align*}
\sum_{b=1}^s(-1)^{1+b}\partial_{x_{b_b}}D_b(v)=0,
\end{align*}
because each mixed second derivative term occurs twice with opposite signs after interchanging the two differentiated columns. Hence
\begin{align*}
M_{A,B}(Jv)=\sum_{b=1}^s(-1)^{1+b}\partial_{x_{b_b}}(v_{a_1}D_b(v)).
\end{align*}
Multiplying by $\varphi$ and integrating by parts over $V$ gives
\begin{align*}
\int_V \varphi(x)M_{A,B}(Jv)(x)\,d\mathcal L^n(x)=-\sum_{b=1}^s(-1)^{1+b}\int_V v_{a_1}(x)\partial_{x_{b_b}}\varphi(x)D_b(v)(x)\,d\mathcal L^n(x),
\end{align*}
with no boundary term because $\operatorname{supp}\varphi\subset V$ is compact. For $v\in W^{1,p}(V;\mathbb R^m)$, this follows by smooth approximation and the multilinear Hölder estimate for minors. Applying this identity with $v=u_k$ and with $v=u$, the tested $s$-minor is a finite sum of exactly the terms whose limits we have just computed. Therefore
\begin{align*}
\lim_{k\to\infty}\int_\Omega \varphi(x)M_{A,B}(Ju_k)(x)\,d\mathcal L^n(x)=\int_\Omega \varphi(x)M_{A,B}(Ju)(x)\,d\mathcal L^n(x).
\end{align*}
This proves distributional convergence for minors of size $s$.
When $p>s$, set
\begin{align*}
\theta_s=\frac{p}{s}.
\end{align*}
Then $\theta_s>1$, so $L^{\theta_s}(\Omega)$ is reflexive by the reflexivity of Lebesgue spaces. The determinant estimate from the integrability step shows that $M_{A,B}(Ju_k)$ is bounded in $L^{\theta_s}(\Omega)$. Let $(M_{A,B}(Ju_{k_\ell}))_{\ell=1}^{\infty}$ be any subsequence. Reflexivity gives a further subsequence, not relabelled, and a function $h\in L^{\theta_s}(\Omega)$ such that
\begin{align*}
M_{A,B}(Ju_{k_\ell})\rightharpoonup h \quad \text{in } L^{\theta_s}(\Omega).
\end{align*}
Weak convergence in $L^{\theta_s}(\Omega)$ implies distributional convergence against every $\varphi\in C_c^\infty(\Omega)$, so the distributional convergence already proved forces $h=M_{A,B}(Ju)$ as a distribution and hence as an element of $L^{\theta_s}(\Omega)$. Thus every subsequence has a further weakly convergent subsequence with the same weak limit $M_{A,B}(Ju)$. If the full sequence did not converge weakly to $M_{A,B}(Ju)$ in $L^{\theta_s}(\Omega)$, some continuous linear functional on $L^{\theta_s}(\Omega)$ would separate a subsequence from that limit by a positive amount, contradicting the further-subsequence conclusion. Therefore
\begin{align*}
M_{A,B}(Ju_k)\rightharpoonup M_{A,B}(Ju) \quad \text{in } L^{p/s}(\Omega).
\end{align*}
This completes the induction step.[/guided]