The $2$-capacity of a set $\Omega$ sitting inside an open set $V \subset \mathbb{R}^n$ is given by $$\text{cap}_2(\Omega, V) = \inf_{u \in C^\infty_0(V), u|_\Omega \equiv 1} \int_V |\nabla u|^2 dx.$$
Now, if we assume that $\Omega$ is deep inside $V$, that is, suppose $\overline{\Omega} \subseteq V$, then do we have $\text{cap}_2(\Omega, V) = \text{cap}_2(\Omega, \mathbb{R}^n)$? Or at least $\text{cap}_2(\Omega, V) \leq C\text{cap}_2(\Omega, \mathbb{R}^n)$, where $C$ is a positive constant?
You have the following result: Let $U$ be another open set with $\bar U \subset V$. Then, there is $C > 0$, such that for all $\Omega \subset U$ we have $$\operatorname{cap}_2(\Omega, V) \le C \, \operatorname{cap}_2(\Omega, \mathbb{R}^n).$$
The proof is by smooth truncation: You take $\varphi \in C_0^\infty(V)$ with $\varphi = 1$ on $U$. If you than have $u \in C_0^\infty(\mathbb{R}^n)$ with $u = 1$ on $\Omega$, $\varphi \, u$ belongs to $C_0^\infty(V)$ and equals $1$ on $\Omega$. And you can bound its norm by a multiple of the norm of $u$. Hence, you get the desired estimate.
By the way: I think your definition of capacity is slightly wrong. Indeed, the set $\mathbb{Q}^n \cap V$ is countable and, thus, should have capacity zero (for $n > 1$). But with your definition, the capacity is $+\infty$.