The answer seems to be negative.
According to this answer, an operator $X\in\operatorname{M}(n,\mathbb{F}_q)$ is semisimple if and only if $X^{q^m}=X$, where $m=\operatorname{lcm}(2,\ldots,n)$.
Now consider the standard realization of the root system of type $\mathsf{A}_n$ inside $\mathbb{Z}^{n+1}$, where the simple roots are $\alpha_i = e_i-e_{i+1}$ for $i=1,\ldots,n$. Then $$\Lambda = \{ (x_1,\ldots,x_{n+1})\in\mathbb{Z}^{n+1} \mid x_1+\ldots+x_{n+1}=0 \},$$ and the action of $W(\mathsf{A}_n)\cong S_{n+1}$ is by permutations of the entries. One possible choice of the Coxeter element $w$ is the long cycle $$w = (1\ \ 2\ \ \ldots\ \ n+1).$$ For $n=3$ one gets $m=6$. Since $\operatorname{ord}(w)=n+1=4$ and $2^6 = 64 \equiv 0 \pmod{4}$, one has $w^{2^m} = \operatorname{id}$, while $w$ act non-trivially on $\Lambda/2\Lambda$.