Suppose that we have a (combinatorial if necessary) model category $M$, and let $F:\Delta^{op}\rightarrow M$ a simplicial object in $M$, such that for any natural number $n$, $F([n])$ is a fibrant object in $M$. 
We define a new object $X= colim\_{n} F([n]) $. Is it true that $X$ is a fibrant object ?



simplicial objects in a model category
No, it is not.


If what you mean by $\rm colim\_n$ is the actual colimit of $F$ as a diagram of shape $\Delta$, then this colimit is isomorphic to the coequalizer of the two maps $F([1]) \rightrightarrows F([0])$. Coequalizers rarely preserve fibrancy, and with a little thought we can think of a counterexample that extends to a simplicial object. Let $M$ be a model category in which coproducts of fibrant objects are fibrant (e.g. simplicial sets), and let $G:D\to M$ be any diagram of fibrant objects whose colimit is not fibrant (such as a cospan $X \leftarrow \Delta[0] \to Y$ in simplicial sets for almost any Kan complexes $X$ and $Y$). Let $F$ be the simplicial bar construction of $G$, with $F([n]) = \coprod\_{d\_0 \to \cdots \to d\_n} G(d\_0)$. Then each $F([n])$ is fibrant by our assumptions on $M$ and $G$, but the colimit of $F$ is the coequalizer of $\coprod\_{d\_0\to d\_1} G(d\_0) \rightrightarrows \coprod\_d G(d)$, which is just the colimit of $G$ (it is the usual computation of colimits in terms of coproducts and coequalizers).


If instead what you mean by $\rm colim\_n$ is actually the geometric realization, then there is an even easier counterexample. Let $M$ be simplicial sets, let $X$ be a simplicial set that is not a Kan complex, and let $F([n]) = X\_n$ regarded as a discrete simplicial set. Discrete simplicial sets are Kan complexes, so each $F([n])$ is fibrant, but the geometric realization of $F$ is just $X$ itself, which is not fibrant.


