Is it true that $q\nmid k$ where $q$ is the characteristic of a ring and $k$ is the order of any unit? 
Let $(R,+,\times)$ be a finite ring with an identity. (Dis)Prove that $q\nmid
 k$ where $q=\text{Char}(R)$ is the characteristic of a ring and $k=\text{Ord}(r)$ is
  the order of a unit $r$.

If $R=\mathbb{Z}/n\mathbb{Z}$, then $q=n$ and $k\mid \varphi(n)$ where $\varphi$ is the Euler function. It holds.
If $R=\mathbb{F}_{p^s}$, then $q=p$ and $k\mid p^s-1$ where $p$ is prime. It holds.
If $R=GR(p^s,r)$, then $q=p^s$ and $k\mid (p^r-1)p^{(s-1)}$, where $p$ is prime. It holds.
I am wondering if it holds for all rings.

No, you can take products of two rings whose unit groups orders divide each others characteristic.


*

*Take $F_4$ has units $C_3$ and $F_3$ has units $C_2$, so $F_4\times F_3$ which has characteristic 6 has units $C_3\times C_2$.


ADDED:
Another thing you can do is take nilpotent extensions to add more units to a given ring, of orders divisible by the characteristic. For example


*

*$F_{p}[x]/(x^p)$, which has a unit $1+x$ of order $p$.

*Hochschild square-zero extension of $R$ by a module $I$ with some element of order divisible by the characteristic of $R$. These extensions have isomorphic copy of $I$ in their unit group as $1+I \subset R^\times\times I$.
ADDED LATER:
For finite commutative rings, being reduced is equivalent to semisimple (cf. Jacobson radical; Semisimple), in other words a product of finite fields (by Artin-Wedderburn), like in our first example.
