As Mariano has pointed out in the comments, over a local ring, projective and free are equivalent, so you are asking whether every finitely generated torsion-free module over a local domain is free.
This is not the case. Indeed, first observe that over any commutative ring $A$, an ideal $I$ is free if and only if it is principal and torsion-free. If $I$ is principal and torsion-free, then the singleton consisting of any generator is a basis for $I$. On the other hand, note that any two distinct elements $x, y \in I$ are linearly dependent, because $y \cdot x - x \cdot y = 0$ is a nontrivial $A$-linear relation.
Hence, it suffices to give an example of a local domain $A$ with an ideal $I$ that is not principal, since $I$ will be torsion-free but not free. Mariano already suggested an example of such a ring in the comments: for $F$ a field, consider the ring $A := F[X, Y]_{\langle X, Y \rangle}$. Then $A$ is a local ring with maximal ideal $M := \langle X, Y \rangle$. But $M$ cannot be generated by a single element; there are many proofs of this, but one is to observe that $M/M^{2}$ is an $A/M \cong F$-vector space of dimension $2$, and so $M$ cannot be principal or else $M/M^{2}$ would be a $1$-dimensional $F$-vector space.
I think it's worth noting that this kind of example is quite common. If $A$ is a Noetherian local ring with unique maximal ideal $M$, then by Krull's principal ideal theorem, the minimum number of generators of $M$ as an $A$-module is bounded below by the Krull dimension of $A$. Examples of local domains with Krull dimension greater than one abound, and give rise to an example in the same way highlighted above. The particular example we gave belongs to an important class of rings you will encounter when studying commutative algebra, namely so-called regular local rings. A Noetherian local ring $A$ with unique maximal ideal $M$ is a regular local ring if the minimal number of generators of $M$ as an $A$-module is equal to the Krull dimension of $A$.
Best Answer
By the structure theorem and the biadditivity of the Hom-functor, it suffices to show this for cyclic torsion modules, which are all of the form $R/(p^n)$. To construct a non-zero homomorphism $R/(p^n) \to R/(p^m)$ we can do the following: first note that $(R/(p^n))/((p)/(p^n)) \cong R/(p)$ by the third isomorphism theorem, so we get a surjective homomorphism $R/(p^n) \to R/(p)$. To construct a non-zero homomorphism $R/(p) \to R/(p^m)$, just take $r+(p) \mapsto rp^{m-1}+(p^m)$ (you need to check this is well-defined and non-zero, I'll leave that to you.) Now if we compose those two maps, we get a non-zero homomorphism $R/(p^n) \to R/(p^m)$