Suppose that such epimorphism exists, and call it $f$.
For fixed $x\in\mathbb{R}$, take $N = \text{Ker}(f)$ and $S = x\mathbb{Q}$.
Apply the second theorem of isomorphisms, that is: $$(SN)/N\cong S/(S\cap N) $$ But $(SN)/N = \{\text{Ker}(f)\}$, so that $S/(S\cap N) = \{ x+S\cap N: x\in S\} = \{ S\cap N \}$, hence $S = S\cap N$. This means that $S\subset N$, from which follows that $$\mathbb{R} = \text{Ker}(f)$$So that $f(x) = 0$ for all $x\in \mathbb{R}$, hence $f$ is not an epimorphism, and contradiction proofs that such epimorphism cannot exists.
Is this proof correct? Did I miss anything? Can it be done easier? Thank you.
Edit: Sorry, I see now that this proof isn't correct. Clearly, $NS$ doesn't have to equal $N$, which I thought is true because of additive and multiplicative notation confusion.
There is an epi $(\mathbb R,+) \rightarrow (\mathbb Q ,+)$. See this question, especially the answer by N.S.