Let $R$ be a commutative ring and $I$ an ideal of $R$.
Definition: a non zero element $ a\in R$ is nilpotent if $\exists n\in\Bbb N$ s.t. $a^n=0$
It is true that if $R/I$ is an integral domain then there cannot be any nilpotent element in $R/I$. But is the other way also true?
On
A ring is called reduced if the only nilpotent element is $0$.
Suppose $R$ is a commutative ring. Just as $R/I$ is an integral domain iff $I$ is a prime ideal, there is an analogous result that $R/I$ is a reduced ring iff $I$ is a radical ideal, that is, $\sqrt{I}=I$, where $\sqrt{I}=\{x\in R\mid x^n\in I \text{ for some } n\in\mathbb N\}$.
It is obvious that "domain implies reduced," but there are obviously rings which are reduced but which are not domains. In such rings the zero ideal serves as a counterexample to "the other way" you are asking about.
Here is a DaRT search for reduced rings that aren't domains. Keep in mind this list may include noncommutative examples.
No. Let $I$ be the ideal $(6)$ in the ring $R = \mathbb{Z}$. More generally, for an ideal $I = (n)$ in $\mathbb{Z}$ where $n \neq 0, 1$, there are nilpotents in $\mathbb{Z}/I$ whenever $(n)$ has some prime power in its factorization, and there are zero-divisors whenever $(n)$ is not prime.