Here is an elementary construction of such a quotient.
Let $X$ be a $T_0$-space and $B_0, B_1, B_2, \ldots$ a countable base for $X$. For any point $x \in X$ define $N(x) = \{i \in \mathbb{N} \mid x \in b_i\}$$N(x) = \{i \in \mathbb{N} \mid x \in B_i\}$, the index set of basic neighborhoods of $x$. Given a sequence $\alpha : \mathbb{N} \to \mathbb{N}$ let $i(\alpha) = \{\alpha(k) \mid k \in \mathbb{N}\}$, the image of the sequence.
The Baire space $\mathbb{N}^\mathbb{N}$ is countably based and $0$-dimensional with the ultrametric $$d(\alpha, \beta) = 2^{-\min_k (\alpha_k \neq \beta_k)}.$$ Let $D$ be the subspace $$D = \{\alpha \in \mathbb{N}^\mathbb{N} \mid \exists x \in X . i(\alpha) = N(x)\},$$ which consists of those sequences that listenumerate the index set of some point in $X$. Note that the point $x$ in the definition of $D$ is unique for a given $\alpha$, if it exists, because $X$ is $T_0$. Define the map $q : D \to X$ by $$q(\alpha) = \text{"the $x$ such that $i(\alpha) = N(x)$"}.$$ It is a basic exercise in topology to verify that $q$ is a quotient map. Thus, every countably based $T_0$-space is the quotient of a $0$-dimensional countably based ultrametric space.
Also note that the map $N : X \to \mathcal{P}(\mathbb{N})$ is an embedding when the codomain is equipped with the Scott topology.