Does $G$ being a subgroup of $GL(n, \mathbb Z/p\mathbb Z)$ for all odd primes $p$ imply it is a subgroup of $GL(n, \mathbb{Z})$?

No, the quaternion group $Q_8$ is a subgroup of ${\rm GL}(2,p)$ for all odd primes $p$, but not of ${\rm GL}(2,{\mathbb Z})$.

Here's one thing we can say. Suppose a finite group $G$ embeds into $GL_n(\mathbb{F}_p)$ for all $p$; actually we can ask for the much weaker condition that $G$ embeds into $GL_n(\prod \mathbb{F}_p)$, and also weaken $G$ to be finitely generated. Then $G$ embeds into $GL_n(K)$ where $K$ is the quotient of $\prod \mathbb{F}_p$ by any of its maximal ideals containing the ideal of elements with finite support; these correspond to ultraproducts $\prod \mathbb{F}_p/U$, and importantly are fields of characteristic $0$.

Since $G$ is finitely generated, this embedding actually takes values in the finitely generated subring $R$ of $K$ generated by the matrix entries of a finite set of generators of $G$. So $G$ embeds into $GL_n(R)$ where $R$ is a finitely generated integral domain of characteristic $0$. This sort of manipulation is a key step in the proof of Malcev's theorem that a finitely generated linear group is residually finite.