Since this question could not be answered at math.stackexchange, I would like to try my luck here now:
Does anyone know an example of a unique factorization domain $R$ that is
(i) not a Dedekind domain (or equivalently, not a principal ideal domain) and
(ii) contains some irreducible element $r \in R$ such that the quotient $R/rR$ is finite?
I am grateful for any suggestions.
My answer https://mathoverflow.net/a/292519/19045 provides an answer to your question as well.
Let $S,k,{\frak m}_1, {\frak m}_2$ be as in that answer, and choose the field $k$ to be a finite field. There I say that $S$ is normal and that ${\frak m}_1$ has height 1, but in fact $S$ is regular (in the sense that its local rings are regular local rings) and ${\frak m}_1$ is principal (as you can see in either Nagata's book or the Stacks tag mentioned in my answer linked above). A regular local ring is a UFD, and a semilocal domain all of whose localizations are UFDs is itself a UFD; hence $S$ is a UFD. Since $S$ is 2-dimensional, it is not a Dedekind domain. Since ${\frak m}_1$ is prime, its generator is irreducible. Since $k$ is finite and $S/{\frak m}_1 \cong k$, the ring $S$ satisfies all the desired conditions, and is moreover Noetherian.
This is an attempt to give a more specific and concrete version of Neil Epstein's answer. The initial version was incorrect, and the current version is incomplete.
Let $k$ be a finite field, and fix a sequence of polynomials $u_i(x)\in k[x]$. For $1\leq n<\infty$ put $$ R_n = k[x,y_0,\dotsc,y_n]/(y_i = x(y_{i+1}+u_i(x))+1), $$ then let $R_\infty$ be the colimit of the rings $R_n$. Because the relations give $y_i$ in terms of $y_{i+1}$ we just have $R_n=k[x,y_n]$ for $n<\infty$, and this is a UFD. We have $y_i=1\pmod{x}$ for all $i$ so $R_\infty/x=k$. On the other hand, we have $y_i=y_0-i\pmod{x-1}$ for all $i$, so $R_\infty/(x-1)=k[y_0]$. From this it follows that the ideal $(x-1,y_0)$ cannot be principal. Thus, if we can prove that $R_\infty$ is a UFD, then we are done.
One can check that if $f$ is irreducible in $R_n$ and does not lie in $R_n.(x,y_n-1)$ then $f$ remains irreducible in $R_{n+1}$.
Initially I had hoped to take $u_i(x)=0$ for all $i$. However, in this case we find that the elements $p_n=(1-x)y_n-1\in R_n$ satisfy $p_n=x\,p_{n+1}$ in $R_{n+1}$, and it follows that $p_0$ cannot be factored as a product of irreducibles in $R_\infty$.
I still think (by comparison with the details of the example mentioned by Neil Epstein) that it should be possible to produce an example by choosing the polynomials $u_k(x)$ appropriately, possibly as $u_k(x)=x^{m_k}$ for some rapidly increasing sequence $m_k$, perhaps $m_k=k!$. The point is that a certain power series defined in terms of the numbers $m_k$ should be transcendental over $k(x)$. However, I have not understood all the details yet.
Edit: disregard this answer, I completely misread the question.
Let $D:=\mathbb{Z}[X]$, and let $S:=D\setminus(2D\cup(3,X)D)$. Let
Take a DVR $V$, and let $R=V[X]$ be the polynomial ring on $V$. Then, $R$ is a unique factorization domain (since $V$ is a UFD) but not a principal ideal domain (since $R$ has dimension 2).
Let $\pi$ be a uniformizer of $V$, and let $r:=\pi X-1$. Then, $r$ is irreducible, and $R/rR$ is isomorphic to $V[1/\pi]$, which is equal to the quotient field of $V$. In particular, you can choose it to have any (infinite) cardinality.
The same works if you take any PID with a finite number of maximal ideals instead of $V$, and then instead of $\pi$ take an element in every maximal ideal.
