Since $W^{k,p}_0$ is a closed subspace of $W^{k,p}$, all these results carry over to $W^{k,p}_0$.
If $v_n \rightharpoonup v$ in $W^{k,p}$ and $v_n \to u$ pointwise a.e., then $u=v$.
As the result ``weak equals pointwise limit'' is hard to track down, here is a sketch of a proof:
Let $(\Omega,\mathcal A,\mu)$ be a $\sigma$-finite measure space.
Assume $v_n \rightharpoonup v$ in $L^p(\Omega)$, $1<p<\infty$, and $v_n \to u$ pointwise $\mu$-a.e. on $\Omega$. Then $v=u$ $\mu$-a.e.
Proof. By Fatou Lemma, $u\in L^p(\Omega)$.
Let $B\subset \Omega$ be bounded. Let $\epsilon>0$. By Egorov's theorem, there is a subset $A\subset B$ such that $v_n \to u$ uniformly on $A$ and $\mu(B\setminus A)<\epsilon$. Let $\phi\in L^\infty(\mu)$. Then $\chi_Bv\in L^q(\mu)$ with $\frac1p+\frac1q=1$ and $1<q<\infty$.
Then
$$
|\int_B(v_n-u)\phi d \mu | = |\int_{A \cup (B\setminus A)} ...| \le \mu(A) \|v_n-u\|_{L^\infty(A)} \|\phi\|_{L^\infty}
+ \epsilon^{1/q} \|v_n-u\|_{L^p(B)} \|\phi\|_{L^\infty}
$$
which implies
$$
\limsup |\int_B(v_n-u)\phi d \mu| \le \epsilon^{1/q} 2M\|\phi\|_{L^\infty}
$$
where $M$ is such that $\|v_n\|_{L^p}, \|u\|_{L^p}\le M$. Since $\epsilon>0$ was arbitrary, this proves $\int_B(v_n-u)\phi d \mu \to0$.
Since $L^\infty(B)$ is dense in $L^q(B)$, this proves $v_n \rightharpoonup u$ in $L^p(B)$. This proves $u=v$ $\mu$-a.e. on $B$. Since $\Omega$ is a countable union of sets of bounded measure, $u=v$ $\mu$-a.e. on $\Omega$ follows. $\square$
I believe, this result is also true for $p=1$ (and $q=\infty)$. Then one has to use the Dunford theorem (see Diestel & Uhl), which is a characterization of weakly convergent sequences in $L^1$.
A proof by Banach-Saks property can be found here:
Weak and pointwise convergence in a $L^2$ space
