Here is what's known about this question:
The problem is hard and requires new ideas. The situation in the $SL(4,Z)$ case is very different and the analogy is misleading.
$\Gamma=SL(3,Z)$ contains no singular semisimple elements (of infinite order). This implies that there cannot be "nontrivial" semisimple RAAGs in $\Gamma$. (I call RAAG trivial if it is a free product of free abelian groups of rank $\le 2$.) One (at least, mine) motivation for looking at nontrivial RAAGs is Serre's question going back to 1977 on coherence of $SL(3,Z)$, which is still open.
If $\Lambda<\Gamma$ is Zariski dense and infinite index, then it cannot contain a lattice in the 3-dimensional Heisenberg group. (This follows from a more general theorem of Benoist and Oh.) This result suggests (but does not quite prove) that $\Gamma$ contains no nontrivial RAAGs.
It is still an interesting problem to embed RAAGs discretely in $SL(3,R)$: Such embeddings are known only for "trivial RAAGs". However, even proving existence of a discrete embedding of $Z^2 *Z$ in $SL(3,R)$ (actually, it embeds in a subgroup $SL(3, Z(1/p))$) is nontrivial. It was done independently few years ago by my student, James Forehand and Grisha Soifer. The proof uses "supersingular embeddings" of $Z^2$ in $SL(3,R)$, i.e., a semisimple embedding where both generators and their product map to singular elements. (This is a bit strange at the first glance, since Tit's ping-pong works best when elements are very proximal, which in $SL_3$ means regular.) These embedding results came as the outcome of unsuccessful attempts to play ping-pong with abelian subgroups of $SL(3,Z)$.
The following papers on this topic are known to be wrong:
S. Wang, Representations of surface groups and right-angled Artin groups in higher rank, Algebr. Geom. Topol 7 (2007), 1099-1117.
(the entire proof I think is hopeless)
and
G. Soifer, Free subgroups of linear groups. Pure Appl. Math. Q. 3 (2007), no. 4, part 1, 987-1003.
more precisely, the proof that $Z^2 *Z$ embeds in $SL(3,Z)$ in this paper is (hopelessly) wrong.