As Jason and Gerald predicted, the answer is yes for Polish $X, Y$.
As Nate observed, we may assume that $D$ is $G_\delta$. We may in fact assume that $D$ is a dense $G_\delta$ by discarding all $\mu$-null open sets from $X$. To avoid trivialities, I will also assume that $X$ has no isolated points. Here is the key fact:
Lemma. If $D \subseteq X$ is dense $G_\delta$ then there is a Borel map $h:X \to D$ such that $\lim_{x\to x_0} h(x) = x_0$ for every $x_0 \in D$.
Supose $D = \bigcap_{n\lt\omega} U_n$, where $(U_n)_{n\lt\omega}$ is a descending sequence of open sets. Any function $h:X \to D$ with the property that if $x \in U_n$ then $d(h(x),x) \lt 1/(n+1)$ will be as required. Since $D$ is dense in $X$ and $X$ has no isolated points, it is always possible to find a suitable $h(x) \in D$. To ensure that $h$ is Borel, fix an enumeration $(d_i)_{i \lt \omega}$ of a countable dense subset of $D$ and, if $x \notin D$, define $h(x)$ to be the first element in this list that matches all the requirements.
Now, if $f:X \to Y$ is Borel and continuous on $D$, then $g = f\circ h$ is a Borel function that agrees with $f$ on $D$ and $$\lim_{x \to x_0} g(x) = f(\lim_{x \to x_0} h(x)) = f(x_0) = g(x_0)$$ for all $x_0 \in D$.