The strict answer to your question is no in general. Take a very special case, $R=k[x,y]$, $M=m$ some maximal ideal of $R$, $N=R$. Then $Ass(M) = Ass(N) = \{(0)\}$, but it is not hard to see $Ext^1(M,N) \cong R/m$, so $Ass(Ext^1(M,N)) = \{m\}$.

However, the general question of understanding the associated primes of Ext is harder and I happen to think about it recently. When the ring is regular, one can get a complete (but complicated) description of the *support* of $Ext^i(M,N)$ based only on the depth of the modules $M,N$ locally at the primes in $Spec(R)$. This was announced by Auslander at the end of his ICM 1962 speech. Sadly enough, the paper he referred to seems to be mysteriously lost.

Shameless plug: Together with Ryo Takahashi, we accidentallly managed to recover Auslander's Theorem 3 from the speech cited above (which is about support of Tor). I am optimistic that his final paragraph can be deciphered in the near future.

anycommutative ring, $M$ is a finitelypresentedmodule, and $N$ isanymodule, then $Ass Hom_R(M,N) = Supp M \cap Ass N$. This is indeed well-known. For instance, it appears as an exercise in the book by Bruns and Herzog. – Neil Epstein Mar 3 '12 at 21:22