This is tiny adaptation of demonstration of the theorem 9.20 from Brezis book: Brezis, Haim Functional analysis, Sobolev spaces and partial differential equations. Universitext. Springer, New York, 2011. xiv+599 pp.
Let $L^p(\Omega)^N=L^p(\Omega)\times...\times L^p(\Omega)$ with the norm $\|u\|^p=\int|u|^p$ where $u=(u_1,...,u_n)$. Because $\Omega$ is bounded, we can use the norm $\|u\|_{1,p}=\|\nabla u\|_p$ in $W_0^{1,p}(\Omega)$. Define $T:W^{1,p}\rightarrow L^p(\Omega)^N$ by $Tu=\nabla u$
Because $T$ is a isometry we have that $T$ is a bijection on $T(W^{1,p})$. Let $S=T^{-1}:T(W^{1,p}_0)\rightarrow W_0^{1,p}$
Note that by using the hypothesis the functional $Gv=\int_\Omega gv$ is a well defined bounded linear functional in $W^{1,p}_0$. Define $P:T(W^{1,p}_0)\rightarrow\mathbb{R}$ by $Pv=GSv$
Because $S$ is continuous and linear, we have that $P$ is a bounded linear functional defined in $T(W^{1,p}_0)$. By using Hahn-Banach we can extend $P$ to a bounded linear functional $\tilde{P}$ defined in $L^{q'}(\Omega)^N$ (because $p>q'$, where $\frac{1}{q}+\frac{1}{q'}=1$).
Therefore we can find $F\in L^{q}(\Omega)^N$ such that $\tilde{P}v=\int Fv$
where $Fv=F_1v_1+...+f_nv_n$. If we take $u\in W_0^{1,p}(\Omega)$ then we have $\int gu=\int F\nabla u$
Please verify if it is correct.