I don't think you asked the question you intended. Even though this question is old, I'll say a bit:
For any fixed $n$ and $p$, there are a finite number of homomorphisms $F_2 \rightarrow SL_n(Z/pZ)$.
Hence, there are only a finite number of possibilities for the trace of a group element as a function of the representation. Therefore, there are many duplicates.
I think a better question is to ask whether there are pairs $(g_1, g_2)$ of non-conjugate elements of $F_2$ such that for all primes $p$ and all homomorphisms $F_2 \rightarrow SL_n(Z/pZ)$ the traces are equal. In this form, you can pass to a limit $p \rightarrow \infty$ and conclude the same trace identity would if nonconjugate elements can be true distinguished by traces for an infinite sequence of $SL_n(Z/pZ)$, they could be distinguished by a homomorphisms to $SL_n(\mathbb C)$. Conversely, if there are no trace identities in $SL_n(\mathbb C)$, then there are no trace identities true in all $SL_n(Z/pZ)$, by finding $Z/pZ$ quotients of the ring of coefficients.
Martin Kassabov has been interested in the latter question, and in discussions he and I have had, we've both come to the opinion that there are probably no trace identitites for $SL_n(C)$ when $n > 2$, but it's not easy to find a proof. One possible strategy is to first characterize all trace identities in $SL_2$, and then construct representations in $SL_3$ where they break down. This is interesting to me in any case because of its meaning in 2 and 3-manifold topology -- the trace identities give collections of distinct elements in $\pi_1$ that are forced to have the equal length in any hyperbolic structure.
A weaker question is whether the characteristic polynomials for representations in $SL_n$ can distinguish conjugacy classes in $F_2$.