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 étale $K$-subalgebra of $D$
of dimension $r$ over $K$ (we calculate in $D\otimes_K K_s$).
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.

`Quadratic and hermitian forms' or in`

The book of involutions'. $\endgroup$1more comment