Let $K$ be a closed convex cone in ${\bf Sym}_n({\mathbb R})$. I assume a generic cone: non void interior, strictly convex. Let
$$K^0=\{ S\in{\bf Sym}_n({\mathbb R})\quad|\quad{\rm Tr}(SH)\ge0,\quad\forall H\in K\}.$$
be its dual. Then $K=(K^0)^0$. If $K=Z^0$ for some conical set $Z$ (that is, $tZ=Z$ for $t>0$), it is necessary that $Z\subset K^0$ and $Z$ contains the extremal lines of $K^0$.

Therefore the cone $K$ has the property that you request if, and only if, the extremal lines of its dual $K^0$ are spanned by matrices of the form $xx^T$.

This happens to be true if $K=K^0={\bf Sym}_n^+$, but it fails in general. Just take any finite set of lines in an open half-space, not all of them being spanned by a rank-one symmetric matrix, take $C$ their convex hull, and choose $K=C^0$. Then $K^0=C$ has an extremal line not spanned by an $xx^T$. Such a $K$ does not share the expected property.