The condition you state is not a necessary condition.  You may find much more about these types of questions in Manin's "Cubic Forms".  

First of all, over a finite field every quadric hypersurface has a rational point by Chevalley's theorem (or you can probably reduce this case to Wedderburn's earlier theorem).  Thus, assume that the field is infinite.  Then, by Bertini's Theorem, for a sufficiently general hyperplane section $C$ of your surface, $C$ is a smooth plane conic.  

For the numerical polynomial $P(n) = n+1$, computing Hilbert polynomials with respect to the restriction to $Q$ of $\mathcal{O}_{\mathbb{P}^3}(1)$, the Hilbert scheme $\text{Hilb}^P_{Q/F}$ is a finite, flat scheme over $C$ of degree $2$.  Of course $C$ is geometrically simply connected.  Thus, either the Hilbert scheme is isomorphic to two disjoint copies of $C$ (the "split" case), or the Hilbert scheme is isomorphic to $C\times_{\text{Spec}\ F}\text{Spec}\ E$ for a degree $2$ separable extension of $F$ (the "non-split case").  

In the split case, $Q$ is isomorphic to the product surface $C\times C$ embedded into $\mathbb{P}^3$ by the complete linear system of the Cartier divisor $\Delta(C)\subset C\times C$. If the surface $Q$ has an $F$-point, then that $F$-point projects to an $F$-point of $C$.  Thus, for every smooth, geometrically integral curve $C$ over $F$ of arithmetic genus $0$ that has no $F$-point, $Q=C\times C$ is an example of a smooth quadric surface that has no $F$-point.  On the other hand, identifying $C$ with either one of the connected components of $\text{Hilb}^P_{Q/F}$, also $Q$ is isomorphic to $C\times \mathbb{P}^1$.