It is not absolutely irreducible: over the algebraic closure, it is the direct sum of $V^{Fr ^i}$ ($0\leq i \leq e-1$), where $Fr : {\mathbb F}_q\rightarrow {\mathbb F}_q$ is the map $x\mapsto x^p$.
I think $e$ being odd or even does not matter: $V$ is irreducible over ${\mathbb F}_p$. The restriction of scalars may be viewed as follows. First, $V$ viewed as a ${\mathbb F}_p$ vector space has dimension $3e$, and one has an embedding ${\mathbb F}_q\subset End_{{\mathbb F}_p}(V)$. The algebra $M_3({\mathbb F}_q)$ is simply the commutant of this copy of ${\mathbb F}_q$ in $End _{{\mathbb F}_p}(V)$ and is spanned as a ${\mathbb F}_p$ vector space (a small check) by elements of $Ad (SL_2({\mathbb F}_q))$.
Suppose $W\subset V$ is irreducible for the adjoint action of $G=SL_2({\mathbb F}_q)$. Then $W$ is stabilised by $M_3({\mathbb F}_q)$; in particular, $W$ is an ${\mathbb F}_q$ subspace of $V$ (now viewed as a ${\mathbb F}_q$ vector space). But $SL_2({\mathbb F}_q)$ acts irreducibly on $V={\mathbb F}_q ^3$, hence $W=V$. [This irreducibility fails of course when $p=2$].
[Edit] Checking that $M_3({\mathbb F}_q)$ is the ${\mathbb F}_p$ span $R$ of $SL_2({\mathbb F}_q)$ matrices involves the following. Consider the diagonal group $$T=\{g=\begin{pmatrix} a ^2 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & a^{-2} \end{pmatrix}\}.$$ Since $a\mapsto a^2$ is a nontrivial character of the mutliplicative group (whose order is the element $-1$ in the field ${\mathbb F}_p$, we see that the ${\mathbb F}_p$ span of the matrices in $T$ contains the matrix $$A=\begin{pmatrix} 0 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 0 \end{pmatrix}.$$
Next, the algebra $R$ contains the ${\mathbb F}_p$ span of matrices $Ad (u)$ where $u$ is upper triangular unipotent in $SL_2({\mathbb F}_q)$. This contains a matrix of the form $$ B=\begin{pmatrix} 1 & 1 & 1 \cr 0 & 1 & 2 \cr 0 & 0 & 1 \end{pmatrix}$$. By forming the commutator $[A,B]$ which lies in the algebra $R$, we have $C=\begin{pmatrix} 0 & -1 & 0 \cr 0 & 0 & 2 \cr 0 & 0 & 0\end{pmatrix}$ in $R$. The commutator $[C,B]=\begin{pmatrix} 0 & 0 & 1 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \end{pmatrix}$ (or a nonzero multiple of this matrix). Hence $E=[B,C]\in R$. Conjugating $E$ by matrices of the diagonal $T$, we get the ${\mathbb F}_q$ line $L_{13}$ through $E=E_{13}$.
The commutator of $L_{13}$ with the lower triangular matrices corresponding to $B$ and $C$, we get the ${\mathbb F}_q$ line $L_{12}$ through $E_{12}$, etc. We get $L_{ij}$ for all $i\neq j$ and whose ${\mathbb F}_p$ algebra span is all of $M_3({\mathbb F}_q)$.
I am hoping to get a more civilised proof, using that the algebra $R$ is semi-simple and that the only semi-simple algebras over ${\mathbb F}_q$ are matrix algebras.