Following the paper of Giusto and Simpson (Located sets and reverse mathematics) 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), which asks whether $f:\mathbb C\rightarrow \mathbb R$ 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
Your question is equivalent to a big open question: is the Mandelbrot set computable?