The conjugacy class of a reflection in an infinite irreducible Coxeter group is always infinite. This follows from a result of mine, which was earlier proved by Kleiner and Pelley in the case of a symmetrizable integer Cartan matrix:

Let $W$ be an infinite irreducible Coxeter group, with generating set $S = \{s_1, s_2, \cdots, s_n\}$. Then the word $s_1 s_2 \cdots s_n s_1 s_2 \cdots s_n \cdots s_1 s_2 \cdots s_n$, consisting of any number of repetitions of $s_1 s_2 \cdots s_n$, is reduced.

Now, recall the following criterion for a word to be reduced: Let $x_1$, $x_2$, ..., $x_N$, be any sequence of elements of $S$.

The word $x_1 x_2 \cdots x_N$ is reduced if and only if the $N$ group elements $x_1$, $x_1 x_2 x_1$, $x_1 x_2 x_3 x_2 x_1$, ... $x_1 x_2 \cdots x_N \cdots x_2 x_1$ are all distinct.

Write $c$ for $s_1 s_2 \cdots s_n$. Then, combining these two results, we see that $s_1$, $c s_1 c^{-1} = s_1 s_2 \cdots s_n s_1 s_n \cdots s_2 s_1$, $c^2 s_1 c^{-2}$, $c^3 s_1 c^{-3}$, etcetera are all distinct. So the conjugacy class of $s_1$ is infinite. Since the ordering of $S$ was arbitrary, every element of $S$ has infinite conjugacy class.

I had earlier read the question as asking whether every conjugacy class in an infinite irreducible Coxeter group was infinite. My answer to that remains below. Thanks to Swiat Gal for pointing out the misreading.

This is not true; the class of a translation in an affine group provides a counterexample.

The simplest such counterexample is the group generated by $a$ and $b$ subject to $a^2=b^2=1$. The conjugacy class of $ab$ is $\{ ab, ba \}$, as you can verify by seeing that $a (ab) a^{-1} = b (ab) b^{-1} = ba$ and $a (ba) a^{-1} = b (ba) b^{-1} = ab$. Geometrically, one can think of this as the group of maps $\mathbb{R} \to \mathbb{R}$ of the form $x \mapsto \pm x + k$, for $k \in \mathbb{Z}$. The generators $a$ and $b$ are $x \mapsto -x$ and $x \mapsto -x+1$.

More generally, every affine Coxeter group is of the form $W_0 \ltimes \mathbb{Z}^r$ for some finite group $W_0$. If you take a group element of the form $(e, v)$ with $v$ a nonzero element of $\mathbb{Z}^r$, its conjugacy class will be $\{ (e, gv): g \in W_0 \}$, which is finite. Geometrically, you should think of this as the group of maps $\mathbb{R}^r \to \mathbb{R}^r$ of the form $x \mapsto gx+v$ where $v$ ranges through $\mathbb{Z}^r$ and $g$ ranges through a finite subgroup $W_0$ of $GL_r$. Then the conjugacy class of a translation $x \mapsto x+v$ is finite.