Let $P$ be the poset $(\partial \Delta[1]) \star (\partial \Delta[1])$ (where $\star$ means "join"). Note that the classifying space of $P$ is $S^1$. Moreover, as a poset, (the nerve of) $P$ is 1-coskeletal.

There is a "suspension" $\Sigma P$ of $P$, like Phil Tosteson suggests, but constructed in a more hands-on way: $\Sigma P$ has

two objects $\{-,+\}$,

4 nondegenerate 1-cells, all going from $-$ to $+$, corresponding to the 4 elements of $P$, and

4 nondegenerate 2-cells corresponding to the 4 1-cells of (the nerve of) $P$. (in each of these one of the 1-faces is degenerate; there's a choice to make of which one -- let's say that the $\partial_0$ face is degenerate)

An exhaustive (but not too bad) search reveals that $\Sigma P$ is 2-coskeletal -- this is essentially because $P$ is 1-coskeletal and has no nontrivial "composable pairs". But clearly the Joyal fibrant replacement of $\Sigma P$ is not 2-coskeletal -- we have $Hom_{\Sigma P}(-,+) \simeq S^1$ which is not essentially discrete.

To be a bit more careful about that last claim, think about it this way. If we apply $\mathfrak C$ to $\Sigma P$, then I think it's pretty clear that we get the simplicial category which I'd also denote $\Sigma P$, with two objects $\{-,+\}$, and with the homspace $Hom(-,+)$ given by (the nerve of) $P$. Since every simplicial set is Joyal-cofibrant and $\mathfrak C$ is left Quillen, we haven't messed up the $\infty$-categorical equivalence class of $\Sigma P$.

Then, a Bergner-fibrant replacement of this simplicial category can be found by simply Kan-fibrantly replacing the homspaces levelwise, and we find that indeed we have an $\infty$-category with two objects $-,+$ and the only nontrivial homspace being $Hom(-,+) \simeq S^1$. This is a model-independent statement, so the Joyal-fibrant replacement of $\Sigma P$ likewise has this property, which shows it's not equivalent to an ordinary 1-category, and hence not equivalent to anything Joyal-fibrant and 2-coskeletal.