Is a finitely generated torsion-free module over a UFD necessarily free?
No. Consider for example the polynomial ring over a field in two variables $R=K[x,y]$. $R$ is a UFD. Now consider the ideal $I=(x,y)$ generated by $x$ and $y$. It is easy to show that $I$ is not a free $R$-module, but it certainly is finitely generated and torsion-free.
No. What we need is that the ring is a principal ideal ring, not just a UFD. A counter example has been just given above. Another generalization of a PID is a Dedekind ring. However, a finitely generated torsion-free module over a Dedekind ring is only projective, but not necessarily free (see here.)