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.
If you know that radical ideals are finite intersections of primes, in particular $\sqrt{(0)}=\mathfrak p_1\cap\cdots\cap\mathfrak p_n$. Let $\mathfrak p$ be a minimal prime. Since $(0)\subseteq\mathfrak p$ we get $\sqrt{(0)}\subseteq\sqrt{\mathfrak p}=\mathfrak p$, that is, $\mathfrak p_1\cap\cdots\cap\mathfrak p_n\subseteq\mathfrak p$. It follows that there exists $\mathfrak p_i\subseteq\mathfrak p$ and since $\mathfrak p$ is minimal we must have $\mathfrak p_i=\mathfrak p$. (In other words, the minimal primes of $A$ are among the primes $\mathfrak p_1,,\dots,\mathfrak p_n$.)
Best Answer
It is worth mentioning that in the question you link to, the ring is assumed to be noetherian. On the other hand, in the non-noetherian case, we have the following:
Example. Let $V$ be a valuation ring. Then, the prime ideals of $V$ are totally ordered, i.e., they look like $$0 \subsetneq \mathfrak{p}_1 \subsetneq \mathfrak{p}_2 \subsetneq \cdots \subsetneq \mathfrak{m}$$ by [BouCA, VI.1.2, Thm. 1]. Thus, every valuation ring has a unique prime ideal (if it exists) of a given height.