Here's the problem: there's no such thing as a continuous probability distribution on the rationals.
The reason for this goes back to the axioms of probability theory, one of which states that if you have a countable collection of disjoint events, the probability that any of them happens is the sum of the probabilities of each happening individually.
So in particular, if you have a random variable $X$ which can take on only countably many different values (that is, a countable range), it is determined entirely by the probabilities $\Pr(X = x)$ for $x$ in the range of $X$.
(In contrast, basically any continuous random variable $Y$ will have $\Pr(Y = y) = 0$ for any $y$, whether $Y$ is uniform or normal or anything in between.)
So any rational random variable is necessarily discrete. In particular, if you try to create even a simple uniform distribution on $\mathbb{Q} \cap [0,1]$, you'll run into issues. Suppose $X$ were some sort of rational uniform r.v. Then for any $c \in \mathbb{Q} \cap (0,1)$ and small $\epsilon>0$, $\Pr(X = c) \leq \Pr(c - \epsilon < x < c+\epsilon) = 2\epsilon$. Taking $\epsilon \to 0^+$, we see that $\Pr(X = c) = 0$ for all $c \in \mathbb{Q} \cap (0,1)$. (In fact, it works for $c = 0, 1$ if you make slight alterations for being on the end of the interval.)
So in fact, by countable additivity, $\Pr(0 \leq X \leq 1) = \sum_{c \in \mathbb{Q} \cap [0,1]} \Pr(X = c) = \sum_{c \in \mathbb{Q} \cap [0,1]} 0 = 0$, yet we're supposed to have this probability be $1$.
No matter what kind of 'continuous' distribution you try to put on $\mathbb{Q}$, you run into the same problem.
But let's say that you don't care about countable additivity, but all you want is to say that $\Pr(A \cup B) = \Pr(A) + \Pr(B)$ when $A,B$ are disjoint. Such finitely-additive probability measures can be studied.
You can even construct uniform and normal distributions on $\mathbb{Q}$. For the former, take a sequence of random variables $U_n$ where $U_n$ is a uniformly random integer between $1$ and $n$, divided by $n$. For the latter, take $B_n$ to be the difference between the number of heads and tails in $n^2$ flips of a fair coin, again divided by $n$.
Now we can't necessarily take the limit of these sequences, but we can take what's called an ultralimit, which is a weird, non-constructive version of a limit that always exists, even when the sequence it's given is completely non-convergent. The ultralimit of the $U_n$ sequence is a uniformly distributed rational random variable on $[0,1]$; likewise, the ultralimit of the $B_n$ is a "standard normal" random variable on the rationals.
Of course, ultralimits have their problems, and the theory of finitely-additive measures has its limits -- namely that it cannot handle the taking of limits due to the lack of the Monotone Convergence Theorem.
In conclusion, you can try to do continuous probability on $\mathbb{Q}$, but beware.
