gr.group theory - Torsion-free and torsionless abelian groups

This question is motivated by my most spectacular answer on MO (:

Let $A$ be a module over $mathbb Z$. $A$ is said to be torsion-free if $na=0$ implies $n=0$ or $a=0$ for any $nin mathbb Z, ain A$. $A$ is torsionless if the map $phi: A to A^{**}$ is injective (here ${}^*$ means $text{Hom}_{mathbb Z}(A,mathbb Z)$).

If $A$ is finite, then torsion-free and torsionless are equivalent. In general, it is not hard to see that being torsionless implies torsion-free. On the other hand, $A=mathbb Q$ is torsion-free but not torsionless since $A^*=0$. But the question and answers quoted above (which shows that for $A=mathbb Z[x]$, $phi$ is an isomorphism) raised the following:

Question: If $A$ is a finite $mathbb Z[x_1,...,x_d]$-module, are being torsion-free and torsionless equivalent?