As YCor indicates, the question in the title is different from the one in the body. Are we assuming the prime $p$ is fixed or not? In both cases the answer is "Yes", but if $p$ is not fixed, there are even stronger results. Below $K$ is always a finitely generated field extension $K$ of $\mathbb{Q}$.
Theorem (Casseles): Given any finite subset $0\neq C\subset K$, there are infinitely many primes $p$ such $K$ embeds in $\mathbb{Q}_p$ and the image of $C$ is in $\mathbb{Z}_p^*$.
See the remark by KConrad (which I was not aware of before, thanks!).
Theorem (Breuillard-Gelander, following Tits): For every infinite subset $C\subset K$ there exists a local field $k$ and an embedding $K$ into $k$ such that the image of $C$ is unbounded.
This is Lemma 2.1 in "A topological Tits alternative". It extends the Tits' lemma cited by YCor in his answer.
I am writing this answer to complete the details in the case where the prime $p$ is fixed. In fact, instead of looking at the extension problem from $\mathbb{Q} \to \mathbb{Q}_p$ and $\mathbb{Q}\to K$ to $K\to k$ for some finite extension $\mathbb{Q}_p\to k$, let us replace $\mathbb{Q}$ by any countable field $K_0$ and $\mathbb{Q}_p$ by any local field $k_0$. Then, by induction, we may assume that $K$ is generated over $K_0$ by a single element $t$. Solving the corresponding extension problem now amounts to choosing a finite field extension $k_0\to k$ and fixing an appropriate element in $k$ as the image of $t$. If $t$ is transcendental, take $k=k_0$ and pick any element in it which is transcendental over the image of $K_0$ (which exists by cardinality considerations) as the image of $t$. If $t$ is algebraic, with minimal polynomial $f$, set $k=k_0[x]/(f)$ and send $t$ to the image of $x$ in it. Thus, indeed, the answer to the problem in the body is "Yes".