Let $f:\mathbb{Q}^n \longrightarrow \mathbb{Q}^n$. We can define what it means for such $f$ to be differentiable: (The differential will be a linear transformation $\mathbb{Q}^n \longrightarrow \mathbb{Q}^n$)
The definition over $\mathbb{R}$ uses the fact it is an ordered field, and the norm structure on $\mathbb{R}^n$. We do not have the standard euclidean norm on $\mathbb{Q}^n$ (we cannot take square roots in $\mathbb{Q}$, and normed spaces are usually defined to be vector spaces over $\mathbb{R}$ or $\mathbb{C}$) but we can use the "rational" $1$-norm instead.
(This is only one option which works "intrinsically", i.e does not require using numbers outside $\mathbb{Q}$, we can use other alternatives of course).
We can also say what it means for $f$ to be continuously differentiable: $f=(f_1,\cdots,f_n)$ , $f_i:\mathbb{Q}^n \longrightarrow \mathbb{Q}$, and the $n$ partial derivatives of $f_i$ are functions $\mathbb{Q}^n \longrightarrow \mathbb{Q}$, and we have a notion of continuity for these creatures.
My question: Assume $f:\mathbb{Q}^n \longrightarrow \mathbb{Q}^n$ is continuously differentiable, and that $Df(x)$ is invertible for some $x\in \mathbb{Q}^n$ . Does the conclusion of the inverse function theorem hold?
This question can be clearly generalized to arbitrary ordered fields. Does this adapted version of the theorem holds anywhere besides $\mathbb{R}$?
(I suspect not since the proofs uses compactness of a closed ball in $\mathbb{R}^n$ or the contraction mapping theorem which assume completeness).
No, this is not true, the basic idea for a counterexample already surfaced in your own comments on the question, it is a function which jumps at irrational points. Let $f:\mathbb{Q} \to \mathbb{Q}$ be defined by $$ f(x) = \begin{cases} x & \text{ for } x \notin (0,\sqrt{2})\\ x+\frac{1}{2^n} & \text{ for } x \in \left(\frac{\sqrt{2}}{n+1}, \frac{\sqrt{2}}{n}\right) \end{cases} $$ Then $f$ is continuously differentiable on $\mathbb{Q}$ with $f'(0)=1$, but the image of any neighborhood of $0$ under $f$ is not a neighborhood of $0$, since it omits all the intervals $\left(\frac{\sqrt{2}}{n}+ \frac{1}{2^n}, \frac{\sqrt{2}}{n} + \frac{1}{2^{n-1}}\right)$ which contain infinitely many rational numbers.
In particular, there is not even a densely defined inverse. Also, by modifying this example (replace $+1/2^n$ by $-1/2^n$) you can make $f$ non-injective on any neighborhood of $0$.