Following [Giusto and Simpson, Located sets and reverse mathematics, J. Symbolic Logic Volume 65, Issue 3 (2000), 1451-1480], the Mandelbrot set $M$ is located if the distance function $f:\mathbb C\rightarrow\mathbb R$, $f(x)=d(x,M)$, exists in the model under consideration, which I assume you take to be the model containing only computable objects.

Such locatedness seems to be the same as Conjecture 4 of [Hertling, Is the Mandelbrot set computable?, Math. Logic Quarterly, 51(1):5-18, 2005]: The function $f:\mathbb C\rightarrow \mathbb R$ is computable.

(Here $f$ is computable if you can compute the value $f(x)$ with any desired precision (in terms of the distance $d$) when you know $x$ with sufficient precision. More precisely, the algorithm that computes $f$ might promise to output a rational interval of length at most $2^{-n}$ containing $f(x)$ once it is told an interval of length $2^{-{g(n)}}$ containing $x$, where $g$ is a computable function on $\mathbb N$ of the algorithm's choosing.)

Now, what happens if we replace $\mathbb C$ by $\mathbb Q[i]$? Could there be a way to compute $f(q)$ using a representation of $q$ as a rational, but nevertheless no way to approximate $f(x)$ given an arbitrary $x$, due to a lack of a useful modulus of continuity in $x\mapsto d(x,M)$?

No, because $|d(x,M)-d(q,M)|\le d(q,x)$.

Conclusion

The question is equivalent to a known-to-be-open question.