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.

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 if $I$ is the annihilator of an element of a free $R$-module, either

• there is some $N$ so that $e_i\in I$ for all $i>N$, or
• there is some $N$ so that $e_i\not\in I$ for all $i>N$.

But the element $x=(e_1,e_3, e_5, \dots)$ of the countable product of copies of $R$ is annihilated by $e_{2i}$ but not by $e_{2i+1}$ for every $i$, and so the submodule generated by $x$ cannot be a submodule of a free module.