Yes. If you are given a simplicial set $X: \Delta^{\text{op}} \to \text{Sets}$, then
the the thick realization $||X||$  of $X$ is given by the same formula as the ordinary realization with the exception that one only uses injective order preserving maps of finite ordered sets. The map $||X|| \to |X|$ is always a homotopy equivalence (here's a reference: https://ncatlab.org/nlab/show/geometric+realization+of+simplicial+topological+spaces#GoodAndProper).

Now, given an ordered simplicial complex $K$,
the promotion $Ss(K)$ of $K$ to a simplicial set
has the property that $||Ss(K)|| = |K|$ identically, where $|K|$ means the geometric realization of $K$ as a simplicial complex. So,
$$
|K| = ||Ss(K)|| \simeq |Ss(K)|\, .
$$.

**Edit:** the second paragraph isn't correct as stated. It isn't true that $||Ss(K)|| = |K|$ identically as Neil points out. I am going to keep this post here until I can find a correction.

**Correction:** If $X$ is a simplicial set, we can consider the poset $P_X$ of its **non-degenerate** simplices. The nerve of this poset is the barycentric subdivision of $X$.  If $X = Ss(K)$, where $K$ is as above, then $P_X$ coincides with the poset of simplices of $K$ under inclusion.

Therefore, with $X= Ss(K)$,  the nerve of the poset $P_X$ coincides with the barycentric subdivision of $K$, so by homotopy invariance of the subdivision, we see that $|X| \simeq |P_X| \simeq |K|$.