Non-UFD integral domain such that prime is equivalent to irreducible?

Yes, e.g. its vacuously true in the ring of all algebraic integers, which has no irreducibles (so no primes), since $\rm\: a = \sqrt{a} \sqrt{a}.\:$ Such domains are known as antimatter domains, since they have no atoms (irreducibles).

If $R$ is a GCD domain, then prime is equivalent to irreducible. Since any Bezout domain is a GCD domain, it is enough to find a Bezout domain which is not an UFD. Such examples can be found here.

A classical example is $R=\mathbb Z+X\mathbb Q[X]$. $R$ is not an UFD since we have the following (strictly) ascending chain of principal ideals: $(X)\subset (\frac{1}{2}X)\subset\cdots\subset(\frac{1}{2^n}X)\subset\cdots$. Note that $R$ is not a "trivial" example, that is, $R$ has irreducible elements, since any prime number is irreducible in $R$. I leave you as an exercise to prove that $R$ is a GCD domain.

Set $\mathbb{Z}_{(2)} = \{\frac{x}{y} \mid x,y\in\mathbb{Z}, 2\nmid y\}$, $R_n = \mathbb{Z}_{(2)}[\sqrt[2^n]{2}]$ and $R = \bigcup_{n\in\mathbb{N}} R_n$. Then $R$ contains no prime elements and no irreducible elements.