Interesting question! Here is a proof when $R$ is a Dedekind domain and $M=R$. The general case for $M$ arbitrary may follow but I have not checked carefully. 

Let $N^*$ denote $Hom(N,R)$ and $D(N)$ denote $Ext_R^1(N,R)$. It is easy to check that for a torsion $R$-module $D^2(N)\cong N$. In particular $ann(D(N)) = ann(N)$.

Now dualize the exact sequence $0 \to I \to R \to R/I \to 0$ one gets:
$$ 0 \to R\to I^* \to D(R/I) \to 0$$

Since $R$ is a Dedekind domain $I^*$ is also an ideal of $R$. It is well-known that ideals in $R$ are strongly 2-generated (they can be generated by at most 2 elements with the first one arbitrary), so $D(R/I)$ is a cyclic $R$-module. But since the annihilator of that module is equal to $I$, it must be $R/I$.   

About the general case, there is a natural map:

$$ Ext^n_R(N,R)\otimes M \to Ext^n_R(N,M) $$
By the case $M=R$ it gives a map $M/IM \to Ext^1_R(R/I,M)$. If one can check that this map localizes naturally, then it reduces to the PID case, but I have not checked. 

Finally, when $R$ is regular local of dimension $n$ then $Ext^n_R(N,R)\cong N$ if $N$ has finite length (by Grothendieck local duality). So perhaps your question can be viewed as some kind of non-local duality.