Actually, no. In any valuation domain, the prime ideals are linearly ordered by inclusion, so there exists at most one nonzero principal prime ideal.
In particular, let $K$ be any field, and let $$R=K[x,y/x,y/x^2,y/x^3,...],$$
i.e., elements of $R$ are "polynomials" in $x$ and $y$ over $K$, except you can divide $y$ by $x$ as many times as you like.
Consider the subset $S$ of $R$ containing all elements with nonzero constant term. It is clear that $S$ is a multiplicative subset of $R$, and let $T=R_S$ be the localization of $R$ at $S$--i.e., $$T=\left\{\frac{f(x,y)}{g(x,y)}\,\vert\,f(x,y)\in R, g(x,y)\in S\right\}.$$
It isn't difficult to show that any nonzero element of $T$ is of the form $ux^n y^m$ where $u\in U(T)$, $m\geq 0$, and if $m=0$ then $n\geq 0$. (Basically, take an arbitrary nonzero element of $T$ and factor out all of the $x$'s and $y$'s that you can.)
So, we have the following chain of prime ideals in $T$ (and, in fact, these are all the prime ideals of $T$): $$0\subsetneq (y,y/x,y/x^2,y/x^3,\cdots)\subsetneq xT$$
You can generalize this to a chain of length $n$ by taking a valuation domain with value group isomorphic to $\mathbb{Z}^n$ under the lexicographic ordering.
Suppose $P$ is of height zero, i.e., a minimal prime of $R$. Then $R_P$, the localization of $R$ at $P$, is a Noetherian local ring of dimension zero, that is, an Artin local ring. The maximal ideal $PR_P$ is therefore nilpotent, so the image of $x$ in $PR_P$ is nilpotent, which means there exists $n\geq 1$, an integer, with $x^n/1=0$ in $R_P$. This means there is some $s\notin P$ with $sx^n=1$. Then $s\neq 0$, and if we choose $n$ minimal, i.e., so that $x^{n-1}/1\neq 0$ in $R_P$, then $sx^{n-1}\neq 0$ but $(sx^{n-1})x=sx^n=0$, so $x$ is a zero divisor, contrary to assumption.
To answer your second question: yes, at least when $R$ is reduced (I'm not sure without this hypothesis). When $R$ is reduced, the set of zero divisors is equal to the union of the minimal primes. Since $R$ is Noetherian, there are finitely many minimal primes, say $P_1,\ldots,P_r$. Now if $Q$ has height one but consists entirely of zero divisors, then $Q\subseteq P_1\cup\cdots\cup P_r$, so by prime avoidance, $Q\subseteq P_i$ for some $i$, a contradiction. So $Q$ must contain a non-zero-divisor.
Best Answer
Noetherian case: Let $p$ be a minimal prime ideal of $R$ and let $0=q_1\cap\cdots \cap q_n$ be a primary decomposition of the zero ideal. Hence $q_1\cap\cdots\cap q_n\subseteq p$ and thus $\sqrt{q_1\cap\cdots\cap q_n}\subseteq p$. Hence we have $\sqrt{q_1}\cap\cdots\cap \sqrt{q_n}\subseteq p$ and so there is an $i$ such that $\sqrt{q_i}\subseteq p$. Now since $p$ is a minimal prime ideal, we have $\sqrt{q_i}=p$. So $p\in \mathrm{Ass}(R)$ and every element of $p$ is a zerodivizor since $Z(R)=\cup_{q\in \mathrm{Ass}(R)} q$.
General case: It is well-known a prime ideal $p$ is minimal prime if and only if for each $x\in p$ there exists $y\in R\setminus p$ such that $xy$ is nilpotent. This shows that every element of a minimal prime ideal is a zero divizor.