This never happens for $q$ odd. If a group $G$ acts on an abelian group $M$, then short exact sequences
$1 \rightarrow M \rightarrow \Gamma \rightarrow G \rightarrow 1$
are classified by elements of $H^2(G;M)$. It is thus enough to show that if $F$ is a field of odd characteristic and if $V = F^n$, then $H^2(GL_n(F);V)=0$.
We have a short exact sequence
$1 \rightarrow \mathbb{Z}/2\mathbb{Z} \rightarrow GL_n(F) \rightarrow GL_n(F)/(\mathbb{Z}/2\mathbb{Z}) \rightarrow 1,$
where $\mathbb{Z}/2\mathbb{Z}$ is generated by the matrix with $-1$'s on the diagonal.
Associated to this is the Hochschild-Serre spectral sequence in cohomology with coefficients in $V$. The $E_2$-term is $H^p(GL_n(F)/(\mathbb{Z}/2\mathbb{Z});H^q(\mathbb{Z}/2\mathbb{Z};V))$. The key fact here is that $H^q(\mathbb{Z}/2\mathbb{Z};V)=0$ for all $q$.
On page 58 of Brown's book on group cohomology, there is a calculation of the cohomology of finite cyclic groups with nontrivial coefficients. In the case we're considering, it goes as follows. Let $t$ be the generator of $\mathbb{Z}/2\mathbb{Z}$. Define $N = 1 + t$. We then get a map $N : V \rightarrow V$ whose image lies in the ring of invariants $V^{\mathbb{Z}/2\mathbb{Z}}$ and which satisfies $N(gv)=N(v)$ for all $g \in \mathbb{Z}/2\mathbb{Z}$ and all $v \in V$. Letting $V_{\mathbb{Z}/2\mathbb{Z}}$ be the ring of coinvariants, we get an induced map $\overline{N}:V_{\mathbb{Z}/2\mathbb{Z}} \rightarrow V^{\mathbb{Z}/2\mathbb{Z}}$. The result then is that $H^0(\mathbb{Z}/2\mathbb{Z};V) = V^{\mathbb{Z}/2\mathbb{Z}}$, that $H^i(\mathbb{Z}/2\mathbb{Z};V) = ker\ \overline{N}$ for $i \geq 1$ odd, and that $H^i(\mathbb{Z}/2\mathbb{Z};V) = coker\ \overline{N}$ for $i \geq 1$ even. But since $F$ has odd characteristic, we have that $V^{\mathbb{Z}/2\mathbb{Z}} = V_{\mathbb{Z}/2\mathbb{Z}} = 0$, and the result follows.