This is essentially the question of whether a $k$-unirational variety is
necessarily $k$-rational.  The short answer is No.
The following longer answer mostly summarizes some of the exposition at
http://en.wikipedia.org/wiki/Rational_variety; for more information
see that page and the references it gives.

The existence of a non-rational subfield $F$ of $K$ depends on $k$ and $n$.
If $k$ is algebraically closed and of characteristic zero,
then the answer is No for $n=2$ by a theorem of
Castelnuovo (and for $n=1$ by a theorem of Lüroth),
but Yes for $n=3$, and thus for all $n \geq 3$
(you did not require $F/K$ to be a finite extension).
In characteristic $p>0$ things can get much stranger:
Zariski gave examples for $n=2$ where the extension $K/F$ is inseparable;
and more recently Shioda constructed, for each $n \geq 2$
and every power $q$ of $p$,
an example where $K/F$ is inseparable and $F$ is the function field of
the Fermat hypersurface of dimension $n$ and degree $q+1$
(which is of general type once $q \geq n+3$), see Propositions
1 and 3 in

> Shioda, T.: An Example of Unirational Surfaces in Characteristic $p$,
> *Math. Ann.* 211 (1974), 233-236.