No.
Let $R$ be the ring of eventually constant sequences of integers, and let $e_i\in R$ be the sequence whose $i^{th}$ term is 1, with all other terms zero.
Then every element of a free $R$-module is annihilated by $e_i-e_{i+1}$ for sufficiently large $i$.
But the element $x=(e_1,e_2,\dots)$ of the countable product of copies of $R$ is not annihilated by $e_i-e_{i+1}$ for any $i$, and so the submodule generated by $x$ cannot be a submodule of a free module.