I would rather write $\chi_{R_1}(a)=\chi_{R_1^*}(a^{-1})$. This allows us to see the sum
$$\frac{1}{n!}\sum_{a\in S_n}\chi_{R_1^*}(a^{-1})\chi_{R_2}(ba)$$ as the trace on $\mathrm{Hom}_{\mathbb{C}}(R_1,R_2)$ of the map $f\mapsto \frac{1}{n!}\sum_{a\in S_n}bafa^{-1}=b\frac{1}{n!}\sum_{a\in S_n}afa^{-1}$. If $b=1$, this is projection to the space of invariants, so its trace is the dimension of this space $\delta_{R_1,R_2}$. If $b\neq 1$, then it's projection to invariants, followed by composition with $b$. If you choose a basis consisting of $b$ (which is sent to $\chi_{R_1}(b)/\chi_{R_1}(1)$ times itself), and then a basis of the other isotypic components in $\mathrm{Hom}_{\mathbb{C}}(R_1,R_2)$ (which are all killed), you can see this gives the factor of $\chi_{R_1}(b)/\chi_{R_1}(1)$ (incidentally, this seems to be a restatement of the proof that this ratio is an algebraic integer).

For the generalization, we still have that $$\frac{1}{n!}\sum_{a\in S_n}\chi_{R_1^*}(a^{-1})\chi_{R_2}(ba)$$ as the trace on $\mathrm{Hom}_{\mathbb{C}}(R_1,R_2)$ of the map $f\mapsto \frac{1}{n!}\sum_{a\in S_n}bafa^{-1}=b\frac{1}{n!}\sum_{a\in S_n}afa^{-1}$. However, this projection might have a much larger image, since there might be many homomorphisms in $\mathrm{Hom}_{S_n}(R_1,R_2)$, since $R_2$ isn't irreducible as an $S_n$-module. My impulse is that there's no easy expression for this trace, but I can't say I'm sure.