Can the following be proven or does there exist a counter example:
The quotient of the ring of integers of a numnber field by an ideal is the direct sum of rings that are either isomorphic to $\mathbb{Z}_m$ or to $\mathbb{F}_q$, where $m$ is an integer and $q$ is the power of a prime.
I don't know if the following outline of a proof is correct. Let $Q$ be the quotient under consideration, and $q_1, \ldots, q_k$ the set of idempotents that are nozero and not the sum of other idempotents (they correspond to the elements of the form $(0,\ldots,1,\ldots,0)$ of Q. We have that $Q=Q.q_1+\ldots+Q.q_k$. Now for a certain $i$ consider the map $Q \rightarrow Q_i=q_1.Q$ that maps $q_1, \ldots, q_k$ to zero except $q_i \mapsto q_i$. The composition of the original quotient map with this map has an ideal as kernel. Either this ideal is maximal (and thus prime) and $Q_i$ is a finite field, either this ideal is not maximal and so $Q_i$ is not a field and the prime ideals of $Q_i$ determine the the prime factorization of the integer $m$.
Marc Bogaerts
A simple counterexample is $\mathbf Z[i]/(2)$. It has size 4, characteristic 2, is not a field, and is not a direct sum of two or more nonzero rings since it is a local ring (i.e., it has one maximal ideal) while the direct sum of two or more nonzero rings is not a local ring.
More generally, suppose for a number field $K$ there is a prime number $p$ such that the ideal $(p)$ in the integers of $K$ is divisible by a prime ideal power $\mathfrak p^r$ where $r > 1$. Then the quotient ring $\mathcal O_K/{\mathfrak p^r}$ is a counterexample: it is not a field since $r>1$ and it has characteristic $p$ since $\mathfrak p^r \mid (p)$, so if it were isomorphic to some $\mathbf Z/(m)$ then that $m$ would have to be $p$, but that would make it a field. It is not isomorphic to a direct sum of the two types of rings you put in your question since it is a local ring, and local rings never decompose as a direct sum of two or more nonzero rings. Every number field other than $\mathbf Q$ has a ramified prime $p$, and (by definition) such a prime is divisible by the square of some prime ideal, so in fact the ring of integers of every number field besides the rationals is a counterexample!