Let $R\subseteq S$ be two integral domains. $S$ is finitely generated over $R$. $Frac(S)$ is finitely generated over $Frac(R)$ (hence, finite by Zariski's lemma). Is it true that $S$ is finite (as module) over some $R[1/f]$ where $0\neq f\in R$?
I have $\mathbb{Z}$ and $\mathbb{Z}[1/2,\sqrt{2}]$ and $\mathbb{Z}$ and $\mathbb{Z}[e]$ in mind. The first can be taken as $\mathbb{Z}[1/2][\sqrt{2}]$ hence finite. Second is not finite but the fraction fields are not finite extension.
This is false even if we assume that the unital homomorphism $R\rightarrow S$ induces an isomorphism on the fraction fields.
The issue is that if $S$ contains $R[1/f]$ for a non-unit $f$ then there is a maximal ideal of $R$ containing $f$ and no proper ideal (in particular, no prime ideal) of $S$ can lie over it. It is not too difficult to produce non-finite ring maps such that every prime ideal of $R$ has a prime ideal of $S$ lying over it. For example, take $R=\mathbb{k}[t^2-1, t^3-t]$, $S=\mathbb{k}[t, \frac{1}{t-1}]$ where $\mathbb{k}$ is an algebraically closed field; the function fields are both $\mathrm{Frac}(\mathbb{k}[t])$.
In the light of this issue I think that the question would be more reasonable if it asks $S\otimes_R R[1/f]$ to be a finite $R[1/f]$-module. I am not sure about this modified question at the moment.