"Is there a property of an ideal $I$ that guarantees that $R/I$ is a principal ideal domain? A Bézout domain? Euclidean?"

First, a clarification. Notice that the condition for an ideal $p$ to be prime is that it is proper (so $p\neq R$) and for any $a,b\in R$, if $ab\in p$, then $a\in p$ or $b\in p$. Thus, when one says that primality is a "property of an ideal", they mean a "property of an ideal inside an intended ring". If you keep $p$ the same, but change $R$, then primality changes. (For instance, if $R=\mathbb{Z}[x]/(3x,x^2)$, and $p=3\mathbb{Z}=3R$, then $p$ is prime as an ideal in $\mathbb{Z}$ but not in $R$.)

So the answer to your questions is yes. Of course there is a property on an ideal $I$ that guarantees that $R/I$ is a PID. You just described it! It is the property that "$R/I$ is a PID".

Perhaps you meant to ask "Is there a **simple** condition on an ideal guaranteeing that $R/I$ has any of those properties?" In the first three examples you gave (of rings that are domains, reduced, or fields), the first two have simple defining conditions in terms of $0$, so they translate cleanly into an ideal-theoretic condition. (In the definition given above of a prime ideal, if we replace "$p$" by the zero ideal, we have exactly the definition of an integral domain.) For fields, they can be characterized by the fact that they have exactly two ideals; it is this alternate definition for fields that translates nicely into ideal-theoretic language.

In the cases of PIDs, Bézout domains, and Euclidean domains, there is no simple defining ring-theoretic condition; and so there is no simple ideal-theoretic condition. (Some people actually study exactly how complicated it is to define these classes of rings, say under the Levy hierarchy of first order logic.)

However, it is commonly the case (especially in noncommutative ring theory) for ring-theoretic conditions (whether simple or not) to have ideal-theoretic analogs. So you are definitely thinking along the right lines! Moreover, sometimes, a simple alternate definition is hiding in plain sight. For example, checking whether an ideal $I$ forces $R/I$ to be reduced is much easier than checking $I=\sqrt{I}$. (For commutative rings you just have to check that if $x^2\in I$, then $x\in I$.)