**first answer.**  
As stated ($X, Y$ merely metric spaces), NO.  
(Remark: we may as well take $D=Y$ and $f$ the identity on $D$: if we can do that case, then we can apply it to get the general case.)  

Let $X=\mathbb R$, let $Y=\mathbb Q$, define $\mu$ a probability measure with positive measure at each rational point of $X$, the rest of $X$ measure zero.  Define $f : \mathbb R \to \mathbb Q$ by $f(x)=x$ for rational $x$ and $f(x)=0$ otherwise.  Then $f$ restricted to $\mathbb Q$ is continuous, and that is a set of measure one.  Suppose $g : \mathbb R \to \mathbb Q$ is an extension of $f$ and every point of $\mathbb Q$ is a point of continuity of $g$.  (A set of measure one must contain all rationals.) As noted in the question (for $\mathbb R$ or even for Polish space, it is presumably older than Kechris), the set $E$ of points where $g$ is continuous is a $G_\delta$.  In particular, since $\mathbb Q$ is not a $G_\delta$ in $\mathbb R$, we see that $E$ contains at least one irrational $u$.  But if $g$ is continuous at $u$ and $g(x)=x$ for all rational $x$, then $g(u)=u$, oops.

Had the second space $Y$ been the completion $\mathbb R$ of $\mathbb Q$, then $g$ would indeed have been a version and a.e. continuous (even continuous everywhere).