Simultaneous similarity of matrices over finite fields

Your intuition is somewhat correct. The characteristic polynomial of $B$ is cubic over $\mathbf F_q$, and has no roots in $\mathbf F_q$ by assumption, hence is irreducible. Thus, the roots all live in $\mathbf F_{q^3}$, and $\mathbf F_q[B] \cong \mathbf F_{q^3}$. If $A$ and $B$ are simultaneously diagonalisable over $\bar {\mathbf F}_q$, then $$\mathbf F_q[A,B] \underset{\mathbf F_q}\otimes \bar{\mathbf F}_q \cong \bar{\mathbf F}_q[PAP^{-1},PBP^{-1}]$$ is a subalgebra of the diagonal matrices $\bar{\mathbf F}_q^3$, so a dimension count shows that the inclusion $\mathbf F_q[B] \subseteq \mathbf F_q[A,B]$ must be an equality. Since $A$ has eigenvalues in $\mathbf F_q$ and $\mathbf F_q[B] \cong \mathbf F_{q^3}$, this forces $A$ constant (think about which elements of $\mathbf F_{q^3}$ have characteristic polynomial totally split over $\mathbf F_q$).

(Conversely, if $A$ is constant, as Will Sawin noted, then clearly $A$ and $B$ are simultaneously diagonalisable over $\bar{\mathbf F}_q$).

I think there is a simpler way to see what is going on: first of all, the hypotheses force the characteristic polynomial of $B$ to be irreducible of degree $3$ over $F_{q}.$ On the other hand, if $PAP^{-1}$ and $PBP^{-1}$ are both diagonal, then $PAP^{-1}$ and $PBP^{-1}$ certainly commute. Hence $A$ and $B$ already commute. Since $B$ has irreducible characteristic polynomial, (which is also its minimum polynomial in this situation), and since $B$ leaves every eigenspace of $A$ invariant (as $A$ and $B$ commute), this forces (given the hypotheses) $A$ to have the form $\lambda I$ for some $\lambda \in F_{q}$.