Misha has said it is an open problem, so I hesitate to say this is an "answer". But, the hypothesis that the rep is faithful means that (after replacing the surface group by a finite index subgroup if necessary) the Zariski closure is connected and not solvable. Hence it has a simple factor, call it $G$. The fact that "all" elements are quasi-unipotent means that the traces (of the adjoint representation of $G$)are all bounded, for all embeddings of the trace field into $\mathbb C$. This is impossible, because that means that the image is finite (this argument is used for example, in Tits' well known paper on the existence of free groups in linear non-solvable by finite groups), and connectedness means that the image is trivial.

Now your hypotheses do not mean all elements have bounded trace, but only that all EMBEDDED loops do; if there is a Zariski dense subset of these embedded loops, we are still OK, I think.