Claim : any rational manifold is a disjoint union of rational ball. (One can probably deduce from this that any two countable manifold of the same dimension are diffeomorphic, but I have a hard time understanding whether or not a ball of transcendental radius is diffeomorphic to $\mathbb{Q}^n$ or not... )
let's prove it for a countable rational manifold:
assume the point of $X$ are numbered $x_1,\dots,x_n...$.
Pick a neighbourhood of $x_1$ that is diffeomorphic to an open ball in $\mathbb{Q}^n$ and then pick a ball arround $x_i$ in that neighbourhood that is both open and closed (and small enough so that it is also closed in $X$) for example, pick a ball whose radius is not a square root of a rational number. Call it $F_1$.
Because $F$ is open and closed, you can easily check that $X$ is diffeomorphic to $(X-F) \cup F$. Do the same in $X - F$ for the next $x_i$ which is not in $F_1$, you get a clopen ball $F_2$ and keep going...
At then end you obtain a partition of $X$ in clopen ball $F_1, \dots F_n \dots...$ and the canonical map between $X$ and the disjoint union of the $F_i$ is a diffeomorphism. (because it is on all of the $F_i$ which form an open cover of $X$).
To break this argument and allow to recover some behavior from real differential geometry what you need is a uniform structure (or a Riemanian/metric structure for example) on your differential manifold compatible with the manifold structure which will force the existence of a "real completion". Or (almost the same) a notion of admissible cover similar to what we have in analytic rigid geometry.