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$is separable. Another version of the proof vas proposed by Uzi Vishne. 1 I answer my own question. The answer is yes. Since there are no non-trivial division algebras over finite fields, we may assume that$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$is a commutative subalgebra of$D$of dimension$r$over$K$(we calculate in$D\otimes_K \bar K$). Clearly$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\$.