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.
See also:
G. Baumslag and J. E. Roseblade, Subgroups of direct products of free groups,
J. London Math. Soc. (2) 30 (1984), 44–52
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
http://front.math.ucdavis.edu/9808.5085
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] http://front.math.ucdavis.edu/9801.5088
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.
