Order of unipotent matrices over $\mathbb{Z}/q\mathbb{Z}$

This works for any unipotent matrix whose characteristic polynomial is $(x-1)^n$ (including some upper-triangular matrices without ones on the diagonal). By Cayley-Hamilton, the matrix $U$ satisfies $(U-1)^{n}=0$, so we can write $U = 1 + N$ with $N^n=0$.

Then $$ U^{p^k} = (1+N)^{p^k} = \sum_{i=0}^{n-1} {p^k \choose i} N^i$$ so it is sufficient to find $k$ such that ${p^k \choose i }=0\mod q$ for all $i$ from $0$ to $n-1$.

Consider the group action of $\mathbb Z/p^k$ by translation on subsets of $\mathbb Z/p^k$ of size $i$. The order of the stabilizer of any point divides $\gcd ( i,p^k)$, so by the orbit-stabilizer theorem, the binomial coefficient is a multiple of $p^k/\gcd(i,p^k)$. The largest possible value of $\gcd(i,p^k)$ is the largest power of $p$ less than $n$, and we want $p^k$ to be $q$ times it, so it suffices to take

$$p^k = q p^ { \lfloor \log_p (n-1) \rfloor }$$

and the order of the matrix is at most this.

One can check this is sharp by taking a single unipotent Jordan block, and checking that the lower bound on the $p$-adic valuation of binomial coefficients is sharp. (At this point one needs to use the fact that the order is necessarily a power of $p$ to reduce to the case of a $p^k$th power.)

Let $q=p^r$. The eigenvalues of a unipotent matrix are all 1 and so we may conjugate it into Jordan canonical form without extending the field. This yields a unipotent upper triangular matrix with entries in ${\Bbb Z}/p{\Bbb Z}$. Thus the answer is independent of $r$.

Consider a Jordan block $A$ of size $m$. Computing its powers we see that its order is $p^t$ where $t=\lceil m / p \rceil$. Thus your answer is $p^{\lceil n / p \rceil}$.