There is no hope for this in any subcanonical topology coarser than the fppf topology, or more generally, any subcanonical topology in which morphisms $\operatorname{Spec} C \to \operatorname{Spec} K$ are not automatically covers when $K$ is a field and $C$ is non-trivial.
(So, as Remy says, this argument does not apply to the fpqc topology.)


Indeed, if $\phi : K \to C$ is any homomorphism that is not a cover and $F$ is the sheaf image of $\operatorname{Spec} \phi : \operatorname{Spec} C \to \operatorname{Spec} K$, then $F$ is not the top subsheaf of $\operatorname{Spec} K$; if we further assume that $A$ is not trivial then $F$ is also not the bottom subsheaf of $\operatorname{Spec} K$.
(We have $\alpha \in F (A)$ if and only if there exist a cover of $\operatorname{Spec} A$ by affines $\operatorname{Spec} \beta_i : \operatorname{Spec} B_i \to \operatorname{Spec} A$ such that each $\beta_i \circ \alpha : K \to B_i$ factors through $\phi : K \to C$.
So $\textrm{id}_K \in F (K)$ if and only if $\operatorname{Spec} \phi : \operatorname{Spec} C \to \operatorname{Spec} K$ is a cover.)

Conversely, by the above argument, in any subcanonical topology such that $\operatorname{Spec} K$ only has the top and bottom subsheaves, it must be that every morphism $\operatorname{Spec} K \to \operatorname{Spec} C$ is either a cover or has $C$ trivial.
But nothing in the argument assumes that $K$ is a field, so it is quite conceivable that even in such a topology, there are non-fields with the property in question.