No. Every ultrafilter $U$ on $\kappa\geq 2$ is a product order, whether it is principal or not. To see this, suppose that $U$ is an ultrafilter on $\kappa\geq 2$.
Partition $\kappa=A\sqcup B$ into two sets with $A\in U$ and $B$ nonempty. 
Every $X\in U$ can be written as
$X=(X\cap A)\sqcup (X\cap B)$, and furthermore, $X\subseteq Y$ just
in case $(X\cap A)\subseteq (Y\cap A)$ and $(X\cap B)\subseteq (Y\cap
B)$. Let $P=U\upharpoonright A=\{ X\subset A\mid X\in U\}$ and
$Q=P(B)=\{X\mid X\subseteq B\}$. These are both nontrivial and $\langle U,\subseteq\rangle$ is
isomorphic to the product order $\langle
P,\subseteq\rangle\times\langle Q,\subseteq\rangle$ by the map $X\mapsto (X\cap A,X\cap B)$.