New answer
The answer is yes.
I think the $\mathbb Q$ and $\overline{\mathbb Q}$ are red herrings.
For instance, if $L/K$ is a field extension and $X$ an indeterminate, one can ask if the inclusion
$$
L(X)\cap K((X))\subset K(X)
$$
holds. (The answer is yes.) This is clearly equivalent to
$$
L(X)\cap K[[X]]\subset K(X).
$$
Now let $f$ be in $K[[X]]$, and let $F$ be $K$ or $L$. Then $f$ is in $F(X)$ if and only if the linear equation
$$
p-fq=0
$$
has a nonzero solution $(p,q)\in F[X]^2$, and we see that the problem makes sense of $K$ and $L$ are commutative rings.
By reasoning like this, we see that the claim boils down to the following statement, which is Corollary $3$ of Proposition II.$3.7$ in Bourbaki's Algèbre.
If $A$ is an associative ring with $1$, if $X$ is a free right $A$-module, and if $(Y_i)_{i\in I}$ is a family of left $A$-modules, then the natural map
$$
\phi:X\ \underset{A}{\otimes}\ \prod_{i\in I}\ Y_i\to\prod_{i\in I}\ \left(X\underset{A}{\otimes} Y_i\right)
$$
is injective.
Proof. Identifying $X$ to a direct sum $\bigoplus_{j\in J}A$ we can write view $\phi$ as a map
$$
\phi:U:=\bigoplus_{j\in J}\ \prod_{i\in I}\ Y_i\ \to\
\prod_{i\in I}\ \bigoplus_{j\in J}\ Y_i=:V
$$
Now $U$ and $V$ are subgroups of
$$
W:=\prod_{(i,j)\in I\times J}Y_i.
$$
More precisely, an element
$$
y=(y_{ij})_{(i,j)\in I\times J}\in W
$$
is in $U$ if and only if there is a finite subset $J(y)$ of $J$ such that $y_{ij}=0$ if $j$ is not in $J(y)$.
Similarly, $y$ is in $V$ if and only if there is, for each $i$ in $I$, a finite subset $J_i(y)$ of $J$ such that $y_{ij}=0$ if $j$ is not in $J_i(y)$.
So we have $U\subset V$, and it is easy to see that $\phi$ is the inclusion.
Old answer
The answer is yes, as follows from the observation below.
Let $A\subset B$ be commutative rings, $x$ an indeterminate, and $f$ an element of $A[[x]]$. Consider the $A[x]$-linear map
$$
\phi:A[x]\times A[x]\to A[[x]],\quad(P,Q)\mapsto P-fQ,
$$
and the $B[x]$-linear map
$$
\psi:B[x]\times B[x]\to B[[x]],\quad(P,Q)\mapsto P-fQ.
$$
If $B$ is $A$-free and $\phi$ injective, then $\psi$ is injective.
Indeed, $\psi$ is the composition of the $B[x]$-linear extension
$$
\phi_B:B[x]\times B[x]\to B\otimes_AA[[x]]
$$
of $\phi$ and the natural map
$$
\theta:B\otimes_AA[[x]]\to B[[x]].
$$
Then $\phi_B$ is injective because $B$ is $A$-flat, and $\theta$ is injective because $B$ is $A$-free.
EDIT A. Assume now that $A$ and $B$ are fields. We claim
$$
B(x)\cap A((x))\subset A(x).
$$
(The reverse inclusion is obvious.) It suffices to show
$$
B(x)\cap A[[x]]\subset A(x).
$$
Let $f$ be in $A[[x]]$. Then $f$ is in $A(x)$ if and only if $\phi$ is not injective, and $f$ is in $B(x)$ if and only if $\psi$ is not injective.
EDIT B. Let me prove the injectivity of $\theta$. Let $(b_i)$ be an $A$-basis of $B$, and let $f$ be in $\ker\theta$. Then $f$ can be written in a unique way as
$$
f=\sum_i\ b_i\otimes f_i
$$
with $f_i\in A[[x]]$ for all $i$, and $f_i=0$ for almost all $i$. Put
$$
f_i=\sum_{n\ge0}\ a_{in}\ x^n
$$
with $a_{in}\in A$. Then we have
$$
0=\theta(f)=\sum_{n\ge0}\ \sum_i\ b_i\ a_{in}\ x^n.
$$
This implies
$$
\sum_i\ b_i\ a_{in}=0
$$
for all $n$, and thus $a_{in}=0$ for all $i$ and all $n$.
EDIT C. Let $A\subset B$ be commutative rings and $x$ an indeterminate. The purpose of this edit is to define $A(x),B(x)$ and $A((x))$ in such a way that
(a) the definitions compatible with the usual ones if $A$ and $B$ are fields,
(b) the equality
$$
B(x)\cap A((x))=A(x).
$$
holds if $B$ free over $A$.
Define $A((x))$ as the ring of Laurent series with coefficients in $A$.
Say that $f\in A((x))$ is in $A(x)$ if there are $P,Q$ in $A[x]$ satisfying $Qf=P$. Then $A(x)$ is a subring of $A((x))$ containing $A[x]$, and the above arguments show that (b) holds.