If $k$ is a field and $V$ is an infinite dimensional $k$-vector space, then $V\cong V\oplus V$. Using this, what you want to show follows from the fact that as $\mathbb Q$-vector spaces, $\mathbb C\cong \mathbb R\oplus \mathbb R$.
Can you see how to prove those two claims?
Later. Ok, apparently not. Let's do it.
(1) If $B$ is a basis for $V$, then the set B'=\{(b,0):b\in B\}\cup \{(0,b):b\in B\} is a basis for $V\oplus V$. There is an obvious bijection B'\cong \{1,2\}\times B.
Now, if $X$ is an infinite set, then $X$ and $\{1,2\}\times X$ are in bijection. It follows from this that there is a bijection between the basis $B$ of $V$ and the basis B' of $V\oplus V$. As you know, this implies that there is a linear isomorphism between $V$ and $V\oplus V$. This proves my first claim above.
(2) Consider the map $\phi:a+bi\in\mathbb C\mapsto (a,b)\in\mathbb R\oplus\mathbb R.$ It is very easy to show that it is an isomorphism of $\mathbb Q$-vector spaces, so that $\mathbb C\cong\mathbb R\oplus\mathbb R$, as my second claim states.
(3) Finally, let's prove your claim that if $\mathbb R$ and $\mathbb C$ are isomorphic $\mathbb Q$-vector spaces: since $\mathbb R$ is a $\mathbb Q$-vector space of infinite dimensions, my first claim tells us that $\mathbb R\cong\mathbb R\oplus\mathbb R$ as $\mathbb Q$-vector spaces. On the other hand, my second claim tells us that $\mathbb R\oplus\mathbb R\cong\mathbb C$ as $\mathbb Q$-vector spaces. Transitivity, then, allows us to conclude that $\mathbb R\cong\mathbb C$ as $\mathbb Q$-vector spaces.