Here's a simple example for $n=4$, namely a finite subgroup of $\mathrm{GL}_4(\mathbf{R})$ not contained in any finite subgroup of $\mathrm{GL}_4(\mathbf{R})$ generated by reflections.
Let $p$ be a prime $\ge 7$, and let $M$ be the $2\times 2$ matrix of order $p$ given by rotation by $2\pi/p$. Consider the group $G\subset\mathrm{GL}_4(\mathbf{R})$ generated by $M_1=\begin{pmatrix}M &0\\ 0 & I_2\end{pmatrix}$, $M_2=\begin{pmatrix}I_2 &0\\ 0 & M\end{pmatrix}$, $s=\begin{pmatrix}0 & I_2\\ I_2 & 0\end{pmatrix}$. (It has order $2p^2$, with an abelian subgroup of index 2 $H$ generated by $M_1,M_2$, and is isomorphic to the wreath product $C_p\wr C_2$ where $C_n$ is cyclic of order $n$.)
I claim that $G$ is not contained in a finite subgroup of $\mathrm{GL}_4(\mathbf{R})$ generated by reflections. There are two steps.
1) $G$ is not contained in a finite subgroup of $\mathrm{GL}_4(\mathbf{R})$ generated by reflexions and normalizing $H$. Indeed, $H\simeq C_p\times C_p$ fixes exactly 2 elements in the 2-Grassmanian of $\mathbf{R}^4$, namely the planes $x_3=x_4=0$ and $x_1=x_2=0$. Hence the normalizer $L$ of $H$ preserves this unordered pair of planes, and hence has a subgroup $L'$ of index (at most) 2 preserving these two subspaces; this index is actually 2 since $s$ belongs to this normalizer but exchanges these 2 subspaces. But $L\smallsetminus L'$ consists of elements of trace zero and hence contains no reflection. Therefore any subgroup of $L$ containing $G$ fails to be generated by reflections.
2) any finite subgroup of $\mathrm{GL}_4(\mathbf{R})$ containing $H$ normalizes $H$. This follows from the classification of subgroups of $\mathrm{SO}(4)$. Indeed, first observe that it follows from the classification of finite subgroups of $\mathrm{SO}(3)$ that in such a finite subgroup, any cyclic subgroup of order $p$ is normal (we use here that $p\neq 2,3,5$), since finite subgroups of $\mathrm{SO}(3)$ with an element of order $p$ are cyclic or dihedral. That cyclic subgroups of order $p$ are normal and $p$-Sylow in all finite subgroups containing them therefore remains true in the double cover $\mathrm{SU}(2)$. It follows that in any finite subgroup $F$ of $\mathrm{SU}(2)\times\mathrm{SU}(2)$, any subgroup of $F$ isomorphic to $C_p\times C_p$ is normal and $p$-Sylow in $F$; and in turn the same conclusion holds in $\mathrm{SO}(4)=(\mathrm{SU}(2)\times\mathrm{SU}(2))/C_2$, and in turn in $\mathrm{O(4)}$, and in turn in $\mathrm{GL}_4(\mathbf{R})$.
The conclusion follows from 1) and 2).
If $B \cong {\mathbb Z}^n$, then the action of $g-1$ on $B$ for $g \in G \setminus \{1\}$ defines an $n \times n$ matrix $A_g$ with entries in ${\mathbb Z}$ and, since $g$ acts semiregularly on $B \setminus \{0\}$, $A_g$ has nonzero determinant.
If you choose your prime $p$ such that it does not divide $\det(A_g)$ for any $g \in G \setminus \{1\}$, then the reduction of $A_g$ mod $p$ will be invertible and so will have zero nullspace. So each $g \in G \setminus \{1\}$ acts semiregularly on the nonzero elements of $B/pB$.
Best Answer
Here is a more direct argument, which avoids using Maschke's Theorem (it also works in infinite-dimensional case).
Let $s, t\in GL(V)$ be involutions fixing the same hyperplane $H\subset V$. Then $s, t$ induce involutions of $V/H\cong F$, where $F$ is your ground field. These induced involutions are either the identity map or the multiplication by $-1$. I will consider the case when both induced involutions are $-1$ since the other cases are ruled out by an argument similar to the one below (we will obtain a contradiction with the assumption that $s, t$ are involutions). Then $g=st$ induces the identity map of $F$ and fixes $H$ elementwise. Choose a vector $v\in V\setminus H$. Then there exists a vector $u\in H$ such that $g(v)=v+u$. Assume $u\ne 0$, otherwise, $g=1$ and $s=t$.
Iterating $g$ we obtain $g^k(v)=v+ ku$, $k\in {\mathbb N}$. Since $F$ has characteristic zero, $ku\ne 0$ for each $k\in {\mathbb N}$. It follows that $s, t$ generate an infinite group, which contradicts our assumptions.