If you really want to define the rationals without using fractions, you could define them as the smallest subset of $\mathbb R$ that contains $1$ and is closed under the four basic arithmetic operations.
(Here we need to explicitly ignore the circular feeling that we need the rationals in order to be sure $\mathbb R$ exists in the first place. If we put our minds to it, we could probably find some way to side-step that -- for example by defining $\mathbb R$ as the completion of the ordered ring of terminating decimal fractions, and then proving separately that what we get is a complete ordered field).
Then we can construct $\mathbb Q$ as follows: Let $f$ be a function that maps finite subsets of $\mathbb R$ to other finite subsets of $\mathbb R$ as follows: $ \begin{align} f(A) = A & \cup \{ a+b \mid a,b\in A \} \cup \{ a-b \mid a,b \in A \} \\& \cup \{ a\times b \mid a,b\in A \} \cup \{ a\div b \mid a,b \in A, b\ne 0 \} \end{align}$ Then by our definition above we can show $ \mathbb Q = f(\{1\}) \cup f(f(\{1\})) \cup \cdots = \bigcup_{n\ge 1} f^n(\{1\}) $ Since a countable union of finite sets is countable, $\mathbb Q$ must be countable.
However, one might very well object that this is really just a thin disguise of choosing arithmetic expressions as "representations" of the rationals.