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".

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 (Cassels): 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 question's body is "Yes".

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 an arbitrary 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 $k=k_0(x)/(f)$ send $t$ to the image of $x$ in it. Thus, indeed, the answer to the problem in the body is "Yes".

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".