Instead of updating my previous answer, I've decided to add a new answer in order to keep it short(ish).

In the comments following his original question, TonyS added the extra assumption that $R$ is finitely generated over $A$. This is a strong condition, since it makes $R$ very close to being commutative. Moreover $R$ is known to be an integral domain, and a maximal order in its division ring of fractions $B$.

Under these assumptions the $A$-torsion-free module $M$ is actually $R$-torsion-free. To see this, let $F$ be the field of fractions of $A$; then since $R$ is finitely generated over $A$, the central localisation $F \otimes_A R$ is an integral domain which is finite dimensional over the field $F$, and is therefore a division ring; thus $F \otimes_A R = B$. Now $F \otimes_A M = F \otimes_A (R \otimes_R M) = (F\otimes_A R) \otimes_R M = B \otimes_R M$ so the kernels of the localisation maps $M \to F \otimes_A M$ and $M \to B \otimes_R M$ coincide. Thus $M$ is $A$-torsion-free if and only if $M$ is $R$-torsion-free.

Now let $N$ be a finitely generated $R$-bimodule which is torsion-free on both sides (for example $N$ could be $\omega_R$). Then $N \otimes_R M$ is a finitely generated $R$-module and therefore a finitely generated $A$-module. To study the torsion $T$ in this module, we study its support $Supp(T)$ in $Spec(A)$, or equivalently, the primes above the annihilator $Ann_A(M)$. Clearly $0$ is not in this support because $T$ is by definition a torsion $A$-module.

I claim that there are no primes in $Supp(T)$ of height $1$. Suppose for a contradiction that $P \in Supp(T)$ has height $1$; then localising $A$ and $R$ at $P$ produces a new maximal order $R_P$ which is free and finitely generated as an $A_P$-module. But $A$ is a commutative regular local ring, hence a UFD by Auslander-Buchsbaum, so $A_P$ is a discrete valuation ring. Since $R_P$ is finitely generated over $A_P$, it must be semilocal; since $R_P$ is also a maximal order, under these conditions it is known that $R_P$ is actually a right and left principal ideal domain: see Proposition 2.9 and Theorem 2.8 of the book "Ordres Maximaux au Sens de K.Asano" by Guy Maury and Jacques Raynaud.

Therefore the module $N_P$ is actually free over $R_P$ and hence $N_P\otimes_{R_P} M_P \cong M_P$ has no torsion. But this module is just $(N \otimes_R M)_P$ and by the exactness of localisation, $T_P$ is a torsion submodule of $(N \otimes_R M)_P$ and is therefore zero: thus $P \notin Supp(T)$, proving the claim.

Now $A$ was assumed to be of dimension at most $2$, so we see that $Supp(T) \subseteq \{ \mathfrak{m} \}$ where $\mathfrak{m}$ is the maximal ideal of $A$. This is the best possible result, because $N \otimes_R M$ can easily have $\mathfrak{m}$-torsion, as the following (commutative!) example shows.

Let $R = A$ and $N = M = \mathfrak{m}$. Pick a regular sequence $x,y$ in $\mathfrak{m}$, so that $0 \to A \stackrel{\alpha}{\to} A^2 \to \mathfrak{m} \to 0$ is a projective resolution of $\mathfrak{m}$, where $\alpha(a) = (ay, -ax)$. Then it's easy to see that $\mathfrak{m} \otimes_A \mathfrak{m} \cong \mathfrak{m}^2 / \alpha(\mathfrak{m})$. Now the image of the element $(y,-x) \in \mathfrak{m}^2$ in $\mathfrak{m}^2 / \alpha(\mathfrak{m})$ is non-zero and is killed by $\mathfrak{m}$, so the $\mathfrak{m}$-torsion submodule of $\mathfrak{m} \otimes_A \mathfrak{m}$ is non-zero.

So to show that $\omega_R \otimes_R M$ has no torsion, one would have to show that $\omega_R$ doesn't "look like" $\mathfrak{m}$ (or perhaps a finite direct sum of copies of $\mathfrak{m}$) as an $A$-module. One way to ensure this is to perhaps try to show that $\omega_R$ is reflexive as an $R$-module, since this would help you to show that there are no essential extensions $E$ of $\omega_R$ such that $E/\omega_R$ is $\mathfrak{m}$-torsion.