$G$ acts irreducibly if $p$ is odd: Suppose that $W$ is an $\mathbb F_p$-subspace of $V$ invariant under $G$, of dimension $d$. So $W$ has $p^d-1$ nonzero elements. Hence there is an orbit of $G$ on these elements of length prime to $p$. Let $A\in W$ be a trace $0$ matrix in this orbit. The orbit length through $A$ is the index $[G:C_G(A)]$. If $A$ has no eigenvalue in $\mathbb F_q$, then $A$ is irreducible on $\mathbb F_q^2$, so by Schur's Lemma we get that $C_G(A)$ is a subgroup of the multiplicative group of $\mathbb F_{q^2}$. In particular, $p$ does not divide $|C_G(A)|$, so $p$ divides the orbit length through $A$, a contradiction.

Thus $A$ has eigenvalues in $\mathbb F_q$. If they are both the same, then they vanish by the trace $0$ condition. If they are however distinct, then $A$ is conjugate in $G$ to a diagonal matrix. So we may assume that there is a nonzero $a$ such that either $A=\begin{pmatrix}0 & a\\0 & 0\end{pmatrix}$ or $A=\begin{pmatrix}a & 0\\0 & -a\end{pmatrix}$. In either case, one quickly computes the conjugacy class $A^G$, and finds out that in both cases they contain an $\mathbb F_p$ basis of $V$.

Haven't checked that case $p=2$, but I expect (in case the conjecture holds) that a similar reasoning should work.

$G$ acts irreducibly on $V$: Suppose that $W$ is an $\mathbb F_p$-subspace of $V$ invariant under $G$, of dimension $d$. So $W$ has $p^d-1$ nonzero elements. Hence there is an orbit of $G$ on these elements of length prime to $p$. Let $A\in W$ be a trace $0$ matrix in this orbit. The orbit length through $A$ is the index $[G:C_G(A)]$. If $A$ has no eigenvalue in $\mathbb F_q$, then $A$ is irreducible on $\mathbb F_q^2$, so by Schur's Lemma we get that $C_G(A)$ is a subgroup of the multiplicative group of $\mathbb F_{q^2}$. In particular, $p$ does not divide $|C_G(A)|$, so $p$ divides the orbit length through $A$, a contradiction.

Thus $A$ has eigenvalues in $\mathbb F_q$. First suppose that $p$ is odd. If the eigenvalues are equal, then they vanish by the trace $0$ condition. If they are however distinct, then $A$ is conjugate in $G$ to a diagonal matrix. So we may assume that there is a nonzero $a$ such that either $A=\begin{pmatrix}0 & a\\0 & 0\end{pmatrix}$ or $A=\begin{pmatrix}a & 0\\0 & -a\end{pmatrix}$. In either case, one quickly computes the conjugacy class $A^G$, and finds out that in both cases the $\mathbb F_p$ span of the classes are $V$. So $W=V$ because $A\in W$.

Similarly, in the case $p=2$ we obtain that we may assume $A=\begin{pmatrix}a & b\\0 & a\end{pmatrix}$ for some $a,b\in\mathbb F_q$. Again, $\mathbb F_p[A^G]=V$.