A finite order element $A\in GL(n,\mathbb{Z})$ has eigenvalues which are roots of unity, and its characteristic polynomial $det(\lambda I-A)$ is a product of cyclotomic polynomials. Thus, it is a necessary condition that the $(\mod p)$ reduction of $A$ have characteristic polynomial which is a $(\mod p)$ reduction of a product of cyclotomic polynomials.

Conversely, suppose that $A_p\in GL(n,\mathbb{Z}/p)$ has a characteristic polynomial which is a $(\mod p)$ reduction of a product of cyclotomic polynomials. Moreover, assume that $A_p$ is conjugate in $GL(n,\mathbb{Z}/p)$ to the companion matrix of its characteristic polynomial (one may check this algorithmically by putting the matrix in rational canonical form over $\mathbb{Z}/p$). Then clearly it is the $(\mod p)$ reduction of a finite order matrix in $GL(n,\mathbb{Z})$. I think this criterion suffices, but I haven't checked it.

A finite order element $A\in GL(n,\mathbb{Z})$ has eigenvalues which are roots of unity, and its characteristic polynomial $det(\lambda I-A)$ is a product of cyclotomic polynomials. Thus, it is a necessary condition that the $(\mod p)$ reduction of $A$ have characteristic polynomial which is a $(\mod p)$ reduction of a product of cyclotomic polynomials.

Conversely, suppose that $A_p\in GL(n,\mathbb{Z}/p)$ has a characteristic polynomial which is a $(\mod p)$ reduction of a product of cyclotomic polynomials. Moreover, assume that $A_p$ is conjugate in $GL(n,\mathbb{Z}/p)$ to the companion matrix of its characteristic polynomial (one may check this algorithmically by putting the matrix in rational canonical form over $\mathbb{Z}/p$). Then clearly it is the $(\mod p)$ reduction of a finite order matrix in $GL(n,\mathbb{Z})$. I think this criterion is necessary and sufficient, but I haven't checked it.

One restriction, the characteristic polynomials of finite-order elements in $GL(n,\mathbb{Z})$ should be products of cyclotomic polynomials. So an element of $GL(n,\mathbb{Z}/p)$ cannot be the image of a finite-order element of $GL(n,\mathbb{Z})$ if its characteristic polynomial is not a $\mod p$ reduction of a degree $n$ polynomial which is a product of cyclotomics.

A finite order element $A\in GL(n,\mathbb{Z})$ has eigenvalues which are roots of unity, and its characteristic polynomial $det(\lambda I-A)$ is a product of cyclotomic polynomials. Thus, it is a necessary condition that the $(\mod p)$ reduction of $A$ have characteristic polynomial which is a $(\mod p)$ reduction of a product of cyclotomic polynomials.

Conversely, suppose that $A_p\in GL(n,\mathbb{Z}/p)$ has a characteristic polynomial which is a $(\mod p)$ reduction of a product of cyclotomic polynomials. Moreover, assume that $A_p$ is conjugate in $GL(n,\mathbb{Z}/p)$ to the companion matrix of its characteristic polynomial (one may check this algorithmically by putting the matrix in rational canonical form over $\mathbb{Z}/p$). Then clearly it is the $(\mod p)$ reduction of a finite order matrix in $GL(n,\mathbb{Z})$. I think this criterion suffices, but I haven't checked it.