I answer my own question. The answer is yes.
Since there are no non-trivial division algebras over finite fields, we may assume that $k$ and $K$ are infinite. Let $H=${$h\in D\ |\ h^\sigma=h$} denote the $k$-space of Hermitian elements of $D$. Consider the embedding $D\hookrightarrow M_r(\bar K)$ induced by an isomorphism $D\otimes_K \bar K\simeq M_r(\bar K)$. An element x of $D$ is called semisimple regular, if its image in $D\otimes_K \bar K\simeq M_r(\bar K)$ is a semisimple matrix that has $r$ different eigenvalues. A standard argument using an isomorphism $D\otimes_k \bar K\simeq M_r(\bar K)\times M_r(\bar K)$ shows that there is a dense Zariski open subset $H_{reg}$ consisting of semisimple regular elements in $H$. Clearly $H_{reg}$ contains $k$-points.
Let $h\in H_{reg}$ be a semisimple regular Hermitian element. Let $L$ be the centralizer of $h$ in $D$. Since $h$ is Hermitian ($\sigma$-invariant), the $k$-algebra $L$ is $\sigma$-invariant. Since $h$ is semisimple and regular, the algebra $L$ is a commutative subalgebra étale $K$-subalgebra of $D$ of dimension $r$ over $K$ (we calculate in $D\otimes_K \bar K$)K_s$). Clearly $L$ is a field, $[L:K]=r$. Since $L\subset D$ and $[L:K]=r$, the field $L$ is a splitting field for $D$, see e.g. Scharlau, Quadratic and Hermitian Forms, Ch. 8, Thm. 5.4.
Since $L\supset K$, we see that $\sigma$ acts non-trivially on $L$. Let $F$ denote the subfield of fixed points of $\sigma$ in $L$, then $[L:F]=2$ and $[F:k]=r$. Clearly $F\cap K=k$ and $FK=L$, hence $L=K\otimes_k F$. The extension $F/k$ is separable.
Another version of the proof vas proposed by Uzi Vishne.
