The answer is almost surely negative. Below are two examples of discrete non-finitely presentable subgroups of $GL(n,{\mathbb C})$ which are generated by finitely many finite order elements. In the first example generators are involutions but some have more than one eigenvalues $-1$, in the second example generators are "complex reflections," of order $5$, but where all but one eigenvalues are equal to $1$. By working harder one should be able to construct examples you are asking for, but I do not see sufficient motivation for doing so.
My guess is that you were hoping for a positive result instead of counter-examples.
Example 1. This example is a variation on the example in
[1] M.Feighn and G. Mess, Conjugacy Classes of Finite Subgroups of Kleinian Groups, American Journal of Mathematics, Vol. 113, No. 1 (1991), 179-188. https://doi.org/10.2307/2374827
See also:
[2] G. Baumslag and J. E. Roseblade, Subgroups of direct products of free groups, J. London Math. Soc. (2) 30 (1984), 44–52, https://doi.org/10.1112/jlms/s2-30.1.44
where it is proven that subgroups of direct products of (virtually) free groups are "almost never" finitely presented.
Let $F$ denote the group with the presentation $\langle x_1, x_2, x_3| x_1^2=x_2^2=x_3^2=1\rangle$. This group can be embedded as a discrete subgroup of
$SU(1,1)$ (the 2-fold cover of $PU(1,1)$, the isometry group of the hyperbolic plane). Next, let $G=F\times F$. Then $G$ embeds discretely in $SU(1,1)\times SU(1,1)\subset SL(4, {\mathbb C})$. Consider homomorphism $\phi: F\to {\mathbb Z}$ sending $x_1, x_2$ to $0$ and $x_3$ to $1$. Let $\Phi=(\phi,\phi): G\to {\mathbb Z}$. Define $\Gamma$ to be the kernel of $\Phi$. Then $\Gamma$ is generated by the following elements:
$$
(x_1,1), (1,x_1), (x_2,1), (1,x_2), (x_3, x_3^{-1}).
$$
Furthermore, $\Gamma$ is not finitely presentable. See [1] where a similar example is explained in detail. Now, every generator of $\Gamma$ is an involution.
Example 2. This is a variation on the example in Michael Kapovich' On normal subgroups in the fundamental groups of complex surfaces https://arxiv.org/abs/math/9808085
Start with the complex-hyperbolic lattice $G$ in $SU(2,1)\subset SL(3, {\mathbb C})$ which appears at number 9 on Thurston's list in [3], page 38.
[3] Shapes of polyhedra and triangulations of the sphere, Geom. Topol. Monogr. 1 (1998), 511-549, https://doi.org/10.2140/gtm.1998.1.511, arXiv:math/9801088
Let $B^2$ be the complex 2-ball on which $G$ acts holomorphically. Geometry of the quotient was understood in detail by Hirzebruch et al. The quotient $B^2/G$, as an orbifold, admits a singular holomorphic fibration whose base is the projective line with three singular points of order $5$, while the generic fiber $\Sigma$ is the projective line with four singular points of order $5$. Now, let $\Gamma\subset G$ be the image of the fundamental group of the generic fiber $\Sigma$ in $G$. The group $\Gamma$ is generated by four complex reflections of order $5$ (each has two eigenvalues $1$ and one
which is 5th root of unity). Then, using the main result of [2], one shows that $\Gamma$ is not finitely presentable. (The proof in [2] applies to torsion-free subgroup of finite index in $\Gamma$, but finite presentability is preserved by passing to finite index subgroups.)
So, what is the difference with the result on real reflection groups? The key is that Tits-Vinberg theorem applies to groups $\Gamma$ generated by real reflections in faces of a convex cone $\Delta$ where "angles'' (whatever this means) have the form $\pi/m_i$. The cone $\Delta$ serves as a fundamental domain for the Coxeter group $\Gamma$ and one can read off the presentation of $\Gamma$ from the combinatorics and geometry of $\Delta$. Such a cone is completely missing in the case of groups generated by complex reflections. Even in the real case, once you drop the "$\pi/m_i$" assumption, the linear representation $\psi$ of the Coxeter group $\Gamma$ (whose presentation is defined by repeating the presentation which appears in the $\pi/m_i$ case) will fail to be faithful. By working harder, one can almost surely get examples of groups $\psi(\Gamma)$ which are not finitely-presented as well.
