I know this is the case if $R$ is a PID, but PID's are special instances of Integral Domains, so I am wondering if there is a counter-example to the case where R is an integral domain.
This post shows the result when $R$ is a Principal Ideal Domain: Proof about finitely generated torsion-free R-module M is free, where R is a PID
Best Answer
Let $R$ be a Dedekind domain, which is not a principal ideal domain, for example the ring of integers in a number field such as $R=\Bbb Z[\sqrt{-6}]$. Then $R$ will have a non-principal ideal, $I$ say. Then, as an $R$-module, $I$ is finitely generated, torsion free, but not principal. In the above example, we could take $I=2R+\sqrt{-6}R$.