Yes, if $m$ generates $M$, it is a basis for $M$, if $R$ is commutative .
Let $b$ be a basis of $M$, so that in particular $Ann(b)=0$.
Since $m$ generates $M$ we can write $b=rm$ for some $r\in R$.
On the other hand we can write $m=sb$ for some $s\in R$ since $b$, a basis, certainly generates $M$.
So we have $b=rm=rsb$, hence $(1-rs)b=0$ and thus $1-rs=0$ because $Ann(b)=0$.
We see that $r,s\in R^*$ are invertible and since $m=sb$ and $b$ is a basis, $m$ is a basis too.
I have used that a basis of a non-zero cyclic free module has just one element.
Since Isaac asks why in a comment, I'll give a proof.
I claim that if $g$ is a generator of $M$, any two elements on $M$ are linearly dependent (still assuming $R$ commutative !)
Indeed, if $u=ag$ and $v=bg$ are arbitrary in $M$, we have a linear relation $bu-av=0$ and either this is a nontrivial linear relation and $u,v$ are linearly dependent or $a=b=0$ and then $u=v=0$ are certainly linearly dependent in that case too.
Important new edit
I had assumed in my proof that $R$ is commutative without saying so. I have now made this assumption explicit: all my apologies to all and thanks to Mariano for calling my attention to this point.