From Lemma 5.2 in Hamenstädt's article Rank-one isometries of proper CAT(0) spaces:
Let $X$ be a proper CAT(0) space and $G \leq \mathrm{Isom}(X)$ a non-elementary group with limit set $\Lambda$ which contains a rank-one element. The action of $G$ on $\Lambda$ is minimal.
If $G$ acts cocompactly on $X$, then $\Lambda= \partial X$.
Therefore, your question reduces to determine whether a group $G$ acting geometrically on a CAT(0) space with isolated flats contains a rank-one element. But an element $g$ that is not rank-one has an axis that bounds a halfplane, so its axis lies in a maximal flat, which has to be stabilised by $g$. Thus, if $g \in G$ is an infinite-order element that does not lie in a maximal abelian subgroup of $G$, then $g$ is rank-one.
(Here, I am assuming that, in the definition of spaces with isolated flats, the space itself is not a flat (up to finite Hausdorff distance).)
Edit: I found a precise reference for the existence of a rank-one element in a group that acts geometrically on a CAT(0) space with isolated flats; see Lemma 2.30 in Cannon-Thurston maps for CAT(0) groups with isolated flats.
In fact, the existence of rank-one elements is tightly connected to hyperbolic properties of the group under consideration. More precisely, given a group $G$ acting on a CAT(0) space $X$ geometrically, we know that $G$ contains a rank-one isometry of $X$ if and only if $G$ is acylindrically hyperbolic. This is essentially due to the equivalence between being rank-one and being Morse; see Theorem 6.40 from the survey arxiv:1709.08843 for more details. Since Hruska and Kleiner proved that groups acting geometrically on CAT(0) spaces with isolated flats are toric relatively hyperbolic and that infinite-order elements that do not lie in parabolic subgroups are Morse, we know that there is plenty of rank-one isometries.