Much stronger statements are true ( and well-known to experts): let $\mathbb{Z}_{p}$ denote the (incomplete) localization at $p$ in $\mathbb{Q}$ ( that is, the rational numbers with denominators prime to $p$ ( together with $0$)).Then when $p$ is odd, only the identity element of ${\rm GL}(n,\mathbb{Z}_{p})$ has $p$-power order and is congruent to the identity (mod $p$). When $p = 2$, any such element has order one or two.

In the case $p$ is odd, we only need to exclude elements of order $p$ which are congruent to the identity (mod $p$). Suppose that $x$ is such an element, and suppose that every entry of $x-I$ is divisible by $p^{j},$ but some element of $x-I$ is not divisible by $p^{j+1}$. Then $x^{p}-I = (x-I)( I + x + \ldots + x^{p-1})$ has every entry divisible by $p^{j+1}$, but
$$I + x + x^{2}+ \ldots + x^{p-1} = pI + (x-I)(I + (x+I) + \ldots + ( I + x + \ldots + x^{p-2}))$$ is congruent to $pI$ ( mod $p^{j+1}$) since $(I + (x+I) + \ldots + ( I + x + \ldots + x^{p-2})$ is conguent (mod $p^{j}$) to $I ( 1 + 2 + \ldots + (p-1)),$ which is congruent to the zero matrix (mod $p$).

Hence some entry of $(x^{p}-I)$ is divisible by $p^{j+1}$, but not by $p^{j+2}$, and certainly $x^{p} \neq I.$ I omit the similar proof when $p=2.$

A similar (well-known) argument proves that if $x \equiv I$ (mod $p^{j})$
( but the congruence fails (mod $p^{j+1}$), and $n$ is not divisible by $p$, then $x^{n}-I \equiv n(x-I)$ ( mod $p^{j+1}$), so $x^{n} \neq I,$ which is sufficient to prove that each element of finite order in $\rm{GL}(n,\mathbb{Z}_{p})$ which is congruent to the identity (mod $p$) has $p$-power order ( so, as seen above, has order $1$ if $p$ is odd, and order dividing $2$ when $p=2$).