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.
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.
Best Answer
Consider the quotient ring $R/P$. $Q/P$ is a prime in $R/P$ of height $n$, hence there exists $\bar x_1, \dots, \bar x_n \in R/P$ such that $Q/P$ is a minimal prime over the ideal $(\bar x_1, \dots, \bar x_n)$. This is equivalent to saying that $Q$ is a minimal prime over $P+(x_1, \dots, x_n)$.
Let me know if this helps!