Let $R$ be a commutative ring and $M$ a finitely generated torsion $R$-module. Then the ideal $\mathrm{Ann}(M)$ is nonzero?
If $R$ is integral domain, this trivially holds since any two ideals have nontrivial intersections.
Add: I say $m\in M$ is torsion, if there is a nonzero element $r\in R$ such that $rm=0$.
With your definition of torsion, this is false. For instance, let $k$ be a field and $R=k[x,y]/(x^2,xy,y^2)$. Let $M$ be the $R$-module with generators $a,b$ and relations $ya=0$, $xb=0$, and $xa=yb$. Note that $M$ is 3-dimensional over $k$, with basis $\{a,b,c\}$ where $c=xa=yb$.
Every element of $M$ is annihilated by some nonzero element of $R$. Indeed, given an element $m=ra+sb+tc$ with $r,s,t\in k$, then $sx-ry$ annihilates $m$ and is nonzero as long as $r$ and $s$ are both not $0$. If $r=s=0$, then $m$ is annihilated by $x$.
However, there is no nonzero element of $R$ that annihilates all of $R$. Indeed, the annihilator of $a$ is $(y)$, the annihilator of $b$ is $(x)$, and the intersection $(x)\cap (y)$ is trivial.