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, but Yes for $n=3$ and presumably for all $n \geq 3$. In positive characteristic things can get much stranger: for each $n \geq 2$ there are inseparable extensions $K/F$ where $F$ is the function field of a variety of general type. Shioda gave explicit examples where $F$ is the function field of the Fermat hypersurface of degree $q+1$ and $q$ is any power of the characteristic; see Propositions 1 and 3 of

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

The exposition and references at http://en.wikipedia.org/wiki/Rational_variety may also be of interest.

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.