Peripheral subgroups of relatively hyperbolic CAT(0) groups are indeed CAT(0) themselves. In fact, more is true: Morse subgroups of CAT(0) groups are CAT(0) themselves.
Definition. Given a finitely generated group $G$, $H \subset G$ is a Morse subgroup if, for every $A>0$ and $B \geq 0$, there exists some constant $K \geq 0$ such that any $(A,B)$-quasigeodesic between any two points of $H$ stays in the $K$-neighborhood of $H$ (in some fixed Cayley graph constructed from a finite generating set; the definition does not depend on this choice).
In his paper Quasi-convexity of hyperbolically embedded subgroups, Sisto proved that hyperbolically embedded subgroups are Morse subgroups, so in particular peripheral subgroups of relatively hyperbolic groups are Morse subgroups.
Now, I claim that Morse subgroups in CAT(0) groups are always convex-compact. (I already proved this statement for groups acting geometrically on CAT(0) cube complexes in my article Hyperbolicities in CAT(0) cube complexes (Proposition 4.2).)
Proposition: Let $X$ be a complete CAT(0) space and $S \subset X$ a Morse subset. The Hausdorff distance between $S$ and its convex hull is finite.
Proof. Let $x$ be a point in the convex hull of $S$, and fix a point $y \in S$. The first observation is that, for every $\epsilon>0$, there exists some $z \in S$ whose projection onto $[x,y]$, say $p$, is at distance less than $\epsilon$ from $x$. Otherwise, if $m$ denotes the point of $[x,y]$ at distance $\epsilon$ from $x$ and $\pi : X \to [x,y]$ the projection onto $[x,y]$, then $\pi^{-1}([m,y])$ would be a convex subspace(*) containing $S$ but not $x$, contradicting the fact that $x$ belongs to the convex hull of $S$.
Now, I claim that $[y,p] \cup [p,z]$ is a $(\sqrt{2},0)$-quasigeodesic. The only point to verify is that, if $a \in [p,y]$ and $b \in [p,z]$, then $d(a,b) \geq \frac{1}{\sqrt{2}} (d(a,p)+d(p,z))$.
Consider a comparison triangle $\Delta = \Delta(\bar{a},\bar{b},\bar{p})$ for $[a,p]\cup[p,b] \cup [a,b]$. Notice that $\angle_{\bar{p}}(\bar{a},\bar{b}) \geq \angle_p(a,b) \geq \pi/2$. Therefore, $d(\bar{a},\bar{b})$ greater or equal to the length of the hypothenuse of a right-angled triangle whose small sides have lengths $\alpha := d(\bar{a},\bar{p})$ and $\beta :=d(\bar{b},\bar{p})$. One has
$$\begin{array}{lcl} d(a,b) & = & d(\bar{a},\bar{b}) \geq \sqrt{\alpha^2+\beta^2} = (\alpha + \beta) \cdot \frac{\sqrt{1+(\alpha/\beta)^2}}{1+ \alpha/\beta} \\ & \geq & \frac{1}{\sqrt{2}} (\alpha+\beta) = \frac{1}{\sqrt{2}} ( d(\bar{a},\bar{p})+d(\bar{b},\bar{p}) \\ & \geq & \frac{1}{\sqrt{2}} ( d(a,p)+d(b,p) \end{array}.$$
Because $S$ is Morse, there exists a universal constant $K$ such that $d(p,S) \leq K$. Consequently, $d(x,S) \leq d(x,p)+d(p,S) \leq K+ \epsilon$.
So the conlusion is: if $K$ is such that any $(\sqrt{2},0)$-quasigeodesic between any two points of $S$ stays in the $K$-neighborhood of $S$, then the Hausdorff distance between $S$ and its convex hull is at most $K$. $\square$
Corollary: Let $G$ be a group acting geometrically on a CAT(0) space $X$. If $H \subset G$ is a Morse subgroup, then $H$ is convex-cocompact, ie., there exists a convex subspace $Y \subset X$ on which $H$ acts cocompactly.
Proof. The map $\left\{ \begin{array}{ccc} G & \to & X \\ g & \mapsto & g \cdot x_0 \end{array} \right.$, where $x_0 \in X$ is a fixed basepoint, is a quasi-isometry, so the orbit $H \cdot x_0$ is a Morse subspace. It follows from the previous proposition that $H$ acts cocompactly on the convex hull of $H \cdot x_0$. $\square$
(*) The assertion is not true. The pre-image of a single point under a projection onto a geodesic is not necessarily convex in a CAT(0) space. So the proof above is not correct. I do not erase the argument in case it would lead to a correct proof.
