In the field of rational numbers, the only valuations are the p-adic valuations. Here is an interpretation of the weak approximation theorem for the field of rational numbers:
Let me say that a (reduced) rational number $q$ is a multiple of $n$ if its numerator is, and its denominator is coprime to n. I note $q \equiv a\ \mod n$ if $q-a$ is multiple of $n$ (one can check that the basic principles of the congruence theory work with this setting, modulo some adaptations).
Now, the weak approximation theorem is equivalent to the following statement:
If $n_1$, $n_2$, ... $n_k$ are coprime natural numbers, and if $q_1$, ... $q_k$ are $k$ rational numbers, then there exists a rational $q$ such that $q \equiv q_i\ \mod n_i$ for every $i$.
It's a "field" version of the Chinese remainder theorem.
But I have some trouble in understanding the meaning of the strong approximation theorem in this setting. Some insight? Do you know a simple interpretation of the strong approximation theorem for $\mathbb Q$?
Well, I asked this question because I had an incomplete statement of the strong approximation theorem; in fact, this is straightforward. The strong approximation theorem simply says that if $S$ is a finite set of primes, then $\mathbb Z$ is dense in $\prod_{p\in S} \mathbb Z_p$ with respect to the p-adic metric, where $\mathbb Z_p$ is the set of p-adic integers.
With the notations in the question, if we take $q_i\in \mathbb Q\cap \mathbb Z_{p_i}$, then $q_i$ is a rational number whose denominator is coprime to $p_i$. The strong approximation theorem then implies that there exists an integer $m$ such that $m \equiv q_i \mod p_i^{n_i}$ for every $i$. That is, the difference with the weak approximation theorem is that $q$ can be chosen to be an integer (but not every possible $q_i$ is allowed, in contrast to the weak approximation theorem).