Assuming by "torsion-free" you mean any zero-divisor on $M$ must be zero-divisor on $R$, which is standard for Noetherian rings. Then you don't need any extra conditions. In fact I will claim the following (no need for flatness of $B$):

Let $(A,m,k)$ be a local, Noetherian, reduced ring. Let $B$ be a Noetherian A-algebra such that $mB \subset rad(B)$ and $M$ be a finitely generated $B$-module. Then $M$ is a flat $A$-module if and only if $M\otimes_Am$ is A-torsion-free.

Proof: One direction is clear, as $m$ is always torsion-free. So now assume $M\otimes_Am$ is A-torsion-free.

Tensor $M$ with $0 \to m \to A \to k\to 0$ we get that $\text{Tor}_1^A(M,k)$ embeds into $M\otimes_Am$. The latter is torsion-free, and the former is killed by $m$. But reducedness implies that $m$ is not an associated prime of $A$ (if it is, then $m$ kills some nonzero element $x\in m$, so $x^2=0$). So $\text{Tor}_1^A(M,k)$ must be $0$, and $M$ is flat by Matsumura Theorem 22.3 (all you need is that $M$ is $m$-adically separated).

Note that without reducedness the result is false. Take $A=B=k[[t]]/t^2$. Then any $A$-module is torsion-free, as everything in $m=(t)$ is a zero-divisor.