Here is a proof that no such examples exist over integral domains:
Let $M$ be finitely generated over an integral domain $R$.
Let $x_1,\dots,x_n$ be generators of $M$. Take $0 \neq a_i \in \operatorname{ann}(x_i)$ for
every $i=1,\dots,n$. Define $a=a_1 \cdots a_n$. Since $R$ is an integral
domain, $a\neq 0$ and $a \in \operatorname{ann}(M)$.
Let me now answer the question negatively in the case where
$R$ is Noetherian and $0 \neq M= J$ is an ideal of $R$. So let us
assume that all elements of $J$ have zero annihilators. Then
all elements of $J$ are zero-divisors of $R$. Let $p_1,\dots,p_s$
be the minimal prime ideals of $R$. We know that the set of
all zero-divisors of $R$ is precisely $p_1 \cup \cdots \cup p_s$.
Consequently, $ J \subset p_1 \cup \cdots \cup p_s$. Then by
prime-avoidance we must have that $J \subset p_1$ without loss of
generality. But the minimal prime ideals have the property that
$p_i = \operatorname{ann}(x_i)$ for some $0 \neq x_i \in R, \, i=1,\dots,s$.
This implies that $J \subset \operatorname{ann}(x_1)$, thus $x_1 \, J=0$ and so
$0 \neq x_1 \in \operatorname{ann}(J)$.