The answer to this question is negative. A suitable counterexample can be constructed as follows.
Let $Y$ be the real line with the starndard Euclidean topology and $X=(\mathbb R\times\{0\})\cup(\mathbb Q\times\{1\})$ be endowed with the topology $\mathcal T$ consisting of the sets $W\subseteq X$ such that for every $(x,k)\in W$ there exist real numbers $a,b$ such that $(x,k)\in \{(y,k):a<y<b,\;y\in S_{x,k}\}\subseteq W$, where $$S_{x,k}=\begin{cases}\mathbb Q&\mbox{if $k=1$};\\ \mathbb R&\mbox{if $k=0$ and $x\in\mathbb Q$};\\ \mathbb R\setminus\mathbb Q&\mbox{if $k=0$ and $x\in\mathbb R\setminus\mathbb Q$}. \end{cases} $$
Therefore, $X$ contains $\mathbb Q\times\{1\}$ as a clopen set and $\mathbb Q\times\{0\}$ as a closed nowhere dense subset.
It is easy to see that the projection $\varphi:X\to Y$, $\varphi:(x,k)\mapsto x$, to the first coordinate is skeletal.
On the other hand, the $\mathbb R$-quotient topology $\tau$ on $Y$ makes the space $(Y,\tau)$ homeomorphic to the subspace $\mathbb R\times\{0\}$ of $X$. In the space $(Y,\tau)$ the set $\mathbb Q$ is closed and nowhere dense. Since $X$ contains $\mathbb Q\times\{1\}$ as a clopen set, the map $\varphi:X\to(Y,\tau)$ is not skeletal anymore.