Every reduction modulo p of an ordinary irreducible lattice stays irreducible

The extension $K\mathrm{/}{\mathbb{Q}}_3$ is totally ramified of degree $2$, with uniformizer $1-{\zeta }_3$, valuation ring $\mathcal{O}$, and residue field $k={\mathbb{F}}_3$. The field $K$ splits the subgroups of $S_3$: it contains the values needed for the cyclic subgroups, and the trivial, sign, and standard representations split $S_3$. If $V$ is a simple $k{S}_3$-module and $g$ generates the normal subgroup $C_3$, then $(g-1{)}^3=g^3-1=0$, so $V^{C_3}\mathrm{\ne }0$; normality and simplicity give $V^{C_3}=V$. Thus $V$ factors through $S_3\mathrm{/}{C}_3\cong C_2$ and is trivial or sign. The same calculation handles the subgroups, so $k$ also splits all of them. Thus the displayed triple is a splitting $3$-modular system, and the free rank-two module $L$ is an $\mathcal{O}{S}_3$-lattice.

The scalar extension $K{\otimes }_{\mathcal{O}}L$ is the standard two-dimensional $S_3$-module. It has no invariant line: a trivial line would be spanned by a constant vector, whose coordinate sum is zero only for the zero vector in characteristic $0$; and a sign line would have to be negated by both $(12)$ and $(23)$, which the coordinate equations again force to be zero. Thus this ordinary representation is irreducible.

By [F1], reduction modulo $\mathfrak{m}=(1-{\zeta }_3)$ gives the two-dimensional $k{S}_3$-module $\bar{L}=L\mathrm{/}\mathfrak{m}L$. The nonzero vector $(1,1,1)$ belongs to $\bar{L}$ because it is the reduction of $(1,1,-2)\in L$, and its line is fixed by $S_3$. Hence $\bar{L}$ is reducible.

This ordinary irreducible lattice has reducible reduction, so the universal statement is false.