This is a corrected answer. I apologize for not posting a complete proof here.
Recall that a group $G$ is called divisible (respectively, uniquely divisible) if for every $g\in G$ and $n\in \mathbb N$, there is $x\in G$ (respectively, there is unique $x\in G$) satisfying $x^n=g$. Recall also that there exist countable (and even finitely generated) torsion free groups where all nontrivial elements are conjugate; we fix one such a group and denote it by $C$. Obviously $C$ is divisible, but not uniquely divisible; moreover, for every prime $p$ and every $g\in C\setminus\{ 1\}$, $g$ has infinitely many $p$th roots.
The following theorem answers the question.
Theorem. There exists a countable uniquely divisible group $G$ and a divisible normal subgroup $H\le G$ such that $G/H$ contains $C$. In particular, there are elements $g\in G/H$ that have infinitely many $p$th roots for every prime $p$.
Unfortunately, I do not know any easy proof. The only proof I know would take few pages. The main idea is to use a modification of the construction from the proof of Theorem 1.5 of my paper http://arxiv.org/abs/math/0411039.