No [Update: well, the answer *was* no to the original question, but the OP later without comment edited the question in a way that invalidates this answer.] . Let $\mu_{p^\infty}$ be the group of $p$th-power roots of unity in $\mathbf C^\times$, with the topology it gets as a subset of $\mathbf C^\times$. This group is not locally compact, since there is no compact neighborhood of 1 in $\mu_{p^\infty}$: a nonempty compact Hausdorff space without isolated points is uncountable, while the full group $\mu_{p^\infty}$ is countable. We'll show $\mu_{p^\infty}$ with its topology inside $\mathbf C^\times$ is the image of $\mathbf Q_p$ under a continuous homomorphism.

Set $\chi \colon \mathbf Q_p \rightarrow \mu_{p^\infty}$ by
$\chi(x) = e^{2\pi i\{x\}_p}$, where $\{x\}_p$ is the $p$-adic fractional part of the $p$-adic number $x$. This $\chi$ is a homomorphism with kernel $\mathbf Z_p$, which is open in $\mathbf Q_p$, so $\chi$ is locally constant on $\mathbf Q_p$ and thus is continuous. It is surjective since for $0 \leq a < p^n$ we have $e^{2\pi i a/p^n} = \chi(a/p^n)$.

Remark: as abstract groups $\mu_{p^\infty} \cong \mathbf Q_p/\mathbf Z_p$, but as topological groups they are not the same since $\mathbf Q_p/\mathbf Z_p$ (with its quotient topology) has the discrete topology. The group $\mathbf Q_p/\mathbf Z_p$ with its discrete topology is locally compact and totally disconnected, so that would not provide a counterexample for your question.