Expanding comment to answer: here's an example that is perhaps somewhat trivial but does achieve what you ask.

Let $(X_0, d, \mu_0)$ be your favorite doubling metric measure space, and suppose $X \subset X_0$ is a subset which is Borel but not $G_\delta$ and has full measure. Consider $(X,d)$ as a metric space in its own right, and equip it with the Borel measure $\mu$ which is the restriction of $\mu_0$ to $X$. (Note that every Borel subset of $(X,d)$ is also a Borel subset of $(X_0, d)$ so this makes sense.) It is a standard theorem that since $X$ is not $G_\delta$ in $X_0$, the metric space $(X,d)$ is not completely metrizable. And for each $x \in X$ and $r \ge 0$, we have $\mu(B_X(x,r)) = \mu_0(B_{X_0}(x, r) \cap X) = \mu_0(B_{X_0}(x,r))$, so $\mu$ is doubling on $X$.

For an explicit example, take $X_0 = [0,1]$, $d$ the usual Euclidean distance, and $\mu_0$ to be Lebesgue measure. For each $n$, let $K_n$ be a fat Cantor set which is compact, nowhere dense, and has $\mu_0(K_n) \ge 1-1/n$. Set $X = \bigcup_n K_n$. Then $X$ is Borel and has full measure; in particular $X$ is dense. But $X$ is also meager, and by the Baire category theorem a dense $G_\delta$ cannot be meager in $[0,1]$. So $X$ is not $G_{\delta}$.

In this case we can actually prove directly that $X$ is not completely metrizable, without needing to appeal to the "standard theorem" mentioned above. We can show that $K_n$ is nowhere dense in $X$, meaning that $X$ is meager in itself, which by Baire category is impossible for a completely metrizable space. Since $K_n$ is compact it is closed in $X$, so we need to show it has empty interior. Let $U$ be nonempty and open in $X$, so that $U = X \cap U_0$ for some nonempty $U_0$ which is open in $X_0 = [0,1]$. Now $K_n$ is closed and nowhere dense in $X_0$, so $U_0 \setminus K_n$ is open and nonempty. Since $X$ is dense, then $\emptyset \ne (U_0 \setminus K_n) \cap X = (U_0 \cap X) \setminus K_n = U \setminus K_n$ so $U$ is not contained in $K_n$. $U$ was arbitrary, so $K_n$ has empty interior and thus is nowhere dense.

One might ask if we can do this for some wide variety of metric measure spaces $(X_0, \mu_0)$. Let's say $(X_0, d)$ is an uncountable Polish space and $\mu_0$ is $\sigma$-finite, Radon, atomless and has full support. Does there necessarily exist a Borel set $X \subset X_0$ which has full measure and is not $G_\delta$? I think a simple cardinality argument will show there must be a *measurable* $X$ having full measure and not $G_\delta$, but we want $X$ to actually be Borel. I am not sure if this is always possible.

**Edit.** The answer to this question is yes. Let $E \subset X_0$ be countable and dense. Since $\mu_0$ is atomless we have $\mu_0(E) = 0$. By regularity of $\mu_0$ there is a $G_\delta$ set $G \supset E$ with $\mu_0(G) = 0$. $G$ is still dense so it is comeager. Set $X = X_0 \setminus G$ so that $X$ has full measure and is Borel (indeed $F_\sigma$). Since $\mu_0$ had full support, $X$ is also dense, but $X$ is also meager so by Baire category it cannot be $G_\delta$.

Moreover, since $G$ is comeager it is uncountable. I believe an uncountable Borel set should contain sets of arbitrarily high Borel rank. By taking complements we can get full-measure Borel sets of arbitrarily high Borel rank.