Over a noetherian ring:

1) the answer is no in general, even for local artinian rings:
Let $R$ be the ring given as a $K$-linear space with basis $(1,t_1,t_2)$ and $t_1t_2=0$, it is local artinian if $K$ is a field.
Let $M$ be the free $K$-module with basis $(e_1,e_2,f)$; make it an $R$-module by setting $t_ie_i=f$, $t_ie_j=0$ for $i\neq j$, $t_if=0$.

It is faithful because if $\alpha+\alpha_1t_1+\alpha_2t_2$ is in the annihilator, evaluation on $f$ yields $\alpha=0$, then on $e_i$ yields $\alpha_i=0$.

But every element has a nontrivial annihilator: an element $a_1e_1+a_2e_2+bf$ is killed by $a_2t_1-a_1t_2$ which is nonzero if $(a_1,a_2)\neq (0,0)$, and $bf$ is killed by $t_1$.

There are many more examples using large abelian subspaces of nilpotent matrices, for instance, using matrices in the upper right block (say over a field) of $2n\times 2n$-matrices, we obtain a $K$-algebra of dimension $n^2+1$ and a faithful $2n$-dimensional module (with dimension in the sense of linear algebra), hence it's to small to contain free submodules of rank 1 as soon as $n\ge 2$.

2)(edit!) the answer is yes if $R$ is an arbitrary domain.

Indeed, let $M$ be a finitely generated $R$-module such that each element has a nonzero annihilator. Using generators, we can write $M$ as a quotient of $\bigoplus_{j=1}^d R/I_j$ for some nonzero ideals $I_j$ (the annihilators of the $d$ generators). Since $R$ is a domain, $I=I_1\dots I_d\neq 0$; since $I$ annihilates $M$ we deduce that $M$ has a nonzero annihilator.