Your proof does not work for the zero ring (fortunately, that is not an integral domain). It has non-prime characteristic $1$.
Apart from that, your proof is ok.
An example is in the integers mod $6$, where $2\cdot 3=0$, but no power of either of these individually is zero.
If we had some sort of measure on a set of rings (which could not possibly be all rings, because there are too many) with finite total measure, we could ask about proportion. I know of no such commonly used measure, but it's possible such a thing has been considered.
In my intuition, the proportion would be small, but I can't think of a quick reason to see why, other than that being nilpotent seems like a special property while being a zero divisor seems very common. For example, the only time this is true for rings of integers modulo $n$ is when $n$ is a prime power, and there are vanishingly few of those compared to all integers.
If you look at the answer by rschwieb in the question Under what conditions does a ring R have the property that every zero divisor is a nilpotent element? linked above in a comment by Dietrich Burde, which is not the accepted answer, it characterizes these rings. A ring has this property if and only if it is the quotient of an arbitrary commutative ring by a primary ideal. An ideal $I$ is said to be primary if whenever we have that $a,b\in R$ satisfy $ab\in I$, then we have either $a\in I$ or $b^n\in I$ for some $n>0$. This answers part of your question. It is nicely illustrated in the $\Bbb Z_n$ case: an ideal $n\Bbb Z$ of $\Bbb Z$ is primary if and only if $n=p^k$ for some prime $p$ and integer $k>0$.
So rings like these are actually "close" to integral domains, which are quotients of arbitrary commutative rings by prime ideals. Again, no quantitative argument for why they should be rare, only an intuitive one.
Best Answer
Suppose $p$ is prime, and let's assume $(a+(p))(b+(p))=(p)$. By the definition of multiplication in the quotient ring it means that $ab+(p)=(p)$, and hence $p|ab$. Since $p$ is prime we conclude that $p|a$ or $p|b$ which implies $a+(p)=(p)$ or $b+(p)=(p)$. So $R/(p)$ has no zero divisors.
For the other direction suppose $R/(p)$ has no zero divisors and assume $p|ab$. Then $(p)=ab+(p)=(a+(p))(b+(p))$. Since there are no zero divisors we conclude that $a+(p)=(p)$ or $b+(p)=(p)$ which means that $p|a$ or $p|b$.