Let $M$ be any module.
If $\{N_i\}_{i\in I}$ is a family of divisible submodules of $M$ (that is, each $N_i$ is a submodule of $M$, and each $N_i$ is divisible), then $N=\langle N_i\rangle$ is also a divisible submodule of $M$: it is a submodulee, and given $n_1+\cdots+n_k\in N$, with $n_j\in N_{i_j}$, and $r\in R$, there exist $m_j\in N_{i_j}$ such that $rm_j = n_j$, and then $r(m_1+\cdots + m_k) = n_1+\cdots + n_k$.
Therefore, if $M$ is any module, then it has a largest divisible submodule (namely, the submodule generated by all divisible submodules; note that the trivial submodule is divisible, so this collection is not empty). Call it $M_d$.
Claim 1. The only divisible submodule of $M/M_d$ is the trivial submodule. that is, $(M/M_d)_d = \{0\}$.
Proof. Let $K$ be a submodule of $M$ that contains $M_d$ such that $K/M_d$ is divisible; let $k\in K$, and let $r\in R$. Then there exists $u+M_d$ such that $r(u+M_d) = (k+M_d)$, so $k = ru+x$ for some $x\in M_d$, with $u\in K$. Since $M_d$ is divisible, there exists $v\in M_d$ (hence in $K$) such that $rv= x$. Thus, $k=r(u+v)$, and $u+v\in K$. Hence $K$ is a divisible submodule of $M$, so $K\subseteq M_d\subseteq K$. That is, $K/M_d$ is the trivial submodule.
Claim 2. $M/M_d$ is reduced.
Proof. Let $A$ be a divisible module, and let $\varphi\colon A\to M/M_d$ be a module homomorphism. Since $\varphi(A)$ is divisible (a quotient of a divisible module is divisible), then by Claim 1 $\varphi(A)=\{0+M_d\}$. Thus, $\varphi$ is the trivial map. Hence, $M/M_d$ is divisible.
This gives you lots of nontrivial reduced modules: just pick any module $M$ that is not divisible, and consider $M/M_d$