There is a rather obvious generalization of your example to cubical pushouts of any dimension. Let $I_n$ be the poset of **proper** subsets of $\{1, ..., n\}$, and $\chi\colon I \to $Sset a diagram indexed on $I$. Let $I_{n-1}$ be the subposet of sets not containing $n$, and $I^n$ the subposet of sets containing $n$ (you can think of these subposets as the back and front faces of a punctured cubical diagram). Suppose that the restrictions of $\chi$ to $I_{n-1}$ and $I^n$ are both cofibrant. Then the map $\mbox{ hocolim }\chi \to \mbox{colim }\chi$ is an equivalence. The reason is that $\mbox{hocolim } \chi$ is equivalent to the homotopy pushout
$$\mbox{hocolim}_{I_{n-1}}\,\, \chi\leftarrow \mbox{hocolim}_{I_{n-1}^1} \,\,\chi \rightarrow \mbox{hocolim}_{I^n} \,\,\chi $$
Here $I_{n-1}^1$ is $I_{n-1}$ minus its final object, which is the set $\{1, \ldots, n-1\}$. There is a similar decomposition of colim $\chi$. It is isomorphic to the strict pushout
$$\mbox{colim}_{I_{n-1}}\,\, \chi\leftarrow \mbox{colim}_{I_{n-1}^1} \,\,\chi \rightarrow \mbox{colim}_{I^n} \,\,\chi $$
Our assumptions guarantee that the left map in this pushout is a cofibration, so pushout=homotopy pushout. The assumptions also guarantee that the natural map from the first pushout diagram to the second is a pointwise equivalence, so it induces an equivalence of homotopy pushouts. So it induces an equivalence hocolim $\chi \to$ colim $\chi$.

Note that $I^n$ is isomorphic to $I_{n-1}$, so by induction you can replace the hypothesis that $\chi$ restricted to ${I^n}$ is cofibrant with a weaker one, so long as $n-1>1$.

The case $n=2$ is equivalent to your example.

Another example: Let $G$ be a finite group (probably can be a more general class of groups - how general?) acting on a **pointed** simplicial set $X$. You can think of this action as defining a diagram of pointed simplicial sets. The colimit of this diagram is the orbit space $X/G$. The homotopy colimit is the pointed homotopy orbit space $X\wedge_G EG_+$. $X$ is cofibrant if the action of $G$ is free except for the basepoint. However, the weaker assumption that $X^H$ is contractible for all non-trivial subgroups $H\subset G$ suffices to conclude that the map from the homotopy orbits space to the strict orbits space is an equivalence.