I believe the following seem to be a very simple counterexample without it being related to Whitehead obstruction as suggested by Tyler Lawson.

Consider the simplicial sets $D$ freely generated by:

It is contractible: indeed it is connected, Its $\pi_1$ is generated by $t$ subject to the relation $(\gamma) : t^2=t$ hence is trivial. Its homology is easy to compute and is trivial.

The inclusion of $\Delta[0]$ into $D$ is hence a cofibration and a weak equivalence, i.e. a trivial cofibration.

$D$ has only three non-degenerate cells, and taking a pushout by a Horn inclusion $\Lambda^k[n] \hookrightarrow \Delta[n]$ always add exactly two non-degenerate cells of dimension $(n-1)$ and $(n)$.

So if $\Delta[0] \hookrightarrow D$ were a pushout of Horn inclusion it would have to be a pushout by $\Lambda^i[2] \hookrightarrow \Delta[2]$.

Now for each $i$ there is only one maps $\Lambda^i[2] \rightarrow \Delta[0]$ along which to take the pushout and none of them gives $D$.

Here is an attempt to get a counter example in the case of the Joyal model structure.

I'm constructing $D'$ by adding to $D$ the following cells:

- a $1$-cell $u:x \rightarrow x$.
- a $2$-cell $\theta$ such that $d_0 \theta =t$, $d_2 \theta = u$ and $d_1 \theta$ is the unique degenerate $1$-cell.

$D'$ is contractible for the Joyal model structure: its homotopy category is freely generated by an idempotent with a left inverse, hence is trivial. Hence $D'$ is an $\infty$-groupoids with trivial $\pi_1$ and the computation of its homology shows that it is contractible.

If it were obtained from $\Delta[0]$ as pushout of Inner Horn inclusion the argument of number of non-degenerate cells shows that it has to be a pushout of to copies of $\Lambda^1[2] \rightarrow \Delta[2]$.

But if that were the case, $t$ would have to be added as pushout of $\Lambda^1[2] \rightarrow \Delta[2]$ at some points, in particular $t$ would appears as $d_1 y$ for $y$ a non-degenerate two cell, with $d_0 y$ and $d_2 y$ not equal to $t$. but there are only two non-degenerate two cells and none of them has this property.

simple homotopy equivalence(more or less by definition). There is an invariant of homotopy equivalences called Whitehead torsion which must vanish for a map $f$ to be homotopic to a simple homotopy equivalence. I'm guessing that you can get a counterexample by taking a homotopy equivalence with nontrivial Whitehead torsion, finding a simplicial model for it, and taking the mapping cylinder to get an inclusion whose realization is a homotopy equivalence not homotopic to a simple one. $\endgroup$ – Tyler Lawson Oct 24 '18 at 10:56