Answer: Yes.
Let $n$ be a positive integer whose value will be determined later on. Let $S$ be the set consisting of $2^n$'s for $i = 0, 1, \dots, 22$. Let $Q$ be the set of subsets of $S$ whose product of own elements is a rational number together with the empty set. Since every positive integer has a unique representation as sum of distinct powers of $2$, the cardinality of $Q$ is precisely the number of integers from $0$ to $K = 2^{23} - 1$ which are divisible by $n$. Therefore,
$$|Q| = 2423 \iff 1 + \left\lfloor \frac{K}{n} \right\rfloor = 2423 \iff 2422 \le \frac{K}{n} < 2423 \iff \frac{K}{2423} < n \le \frac{K}{2422}.$$
Thus, if $K/2422 - K/2423 \ge 1$, then there exists an integer $n$ such that $|Q| = 2423$. As $2422 \cdot 2423 < (2^{11}\sqrt{2} - 1)^2 < 2^{23} - 1 = K$, there is such an $n$. (Indeed, $n = 3463$ fulfills the condition.)