I think it is not necessarily the case. So far I have that if $S\subset R$ is a an integral infinite ring extension, we would need to have that $R$ is infinitely generated but $S[R]$ is finitely generated, both as S-modules.
Im having trouble coming up with a concrete example, though.
Fact: An $S$-Algebra $R$ is finite (i.e. finitely generated as an $S$-module) if and only if it is integral and of finite type (i.e. finitely generated as an $S$-algebra). [I guess you know this but I am not entirely sure whether I understood your notation correctly.]
Hint (for finding the desired example): Take a $S$-algebra $R$ that is not of finite type (and is reasonably easy to handle). Then "force" $R$ to be integral by modding out suitable relations while retaining the property of not being of finite type over $S$.
Edit: I guess the example provided by Timbuc in the comment to the original question is even simpler.