Let $\mathbf{C}$ be the complex numbers, and $\mathbf{Q}_p$ the $p$-adic rationals, embedded into $\mathbf{C}$ by some field embedding $s$.
Is the abelian group $\Gamma := \mathbf{C}/s(\mathbf{Q}_p)$ a torsion abelian group?
$s$ must restrict to the identity on $\mathbf{Q}\subset\mathbf{Q}_p$, so there is a surjective group homomorphism $$\mathbf{C}/\mathbf{Q}=\mathbf{R}/\mathbf{Q}\oplus\mathbf{R}\to\Gamma.$$
Is it true that $s(\mathbf{Q}_p)$ is always not contained in the copy of $\mathbf{R}$ as a subfield of $\mathbf{C}$? If so, the intersection of $s(\mathbf{Q}_p)$ and the second copy of $\mathbf{R}$ in $\mathbf{C} = \mathbf{R}\oplus\mathbf{R}$ as an abelian group, must contain $\mathbf{Z}$, so the surjection factors through $\mathbf{C}/\mathbf{Q}\oplus\mathbf{R}/\mathbf{Z}$, and $\Gamma$ is torsion.
No: in fact, $\Gamma$ must be torsion-free. A very quick way to see this is to observe that $s(\mathbb{Q}_p)$ is a $\mathbb{Q}$-vector subspace of $\mathbb{C}$, so the quotient is again a $\mathbb{Q}$-vector space.
Note that your argument is incorrect: $\mathbb{C}/\mathbb{Q}\oplus\mathbb{R}/\mathbb{Z}$ is not a torsion group! The summand $\mathbb{R}/\mathbb{Z}$ does have some torsion elements, but not all of its elements are torsion (think about irrational numbers), and it's entirely possible that the torsion elements die when you take a further quotient.