Let's say that an $(\infty,1)$-topos is Boolean if for every object $X$, the lattice $Sub(X)$ of subobjects (i.e. $(-1)$-truncated morphisms into $X$) is a Boolean algebra. I think this is equivalent to asking that the subobject classifier be an internal Boolean algebra, and hence to asking that the underlying 1-topos of 0-truncated objects is Boolean in the classical sense.
In particular, the $(\infty,1)$-topos of sheaves on a topological space $X$ is Boolean if and only if the lattice of open sets in $X$ is a Boolean algebra, i.e. every open set is also closed. Thus, Booleanness is a sort of "zero-dimensionality" condition. Lurie shows in Higher Topos Theory that other sorts of "finite-dimensionality" conditions imply that an $(\infty,1)$-topos is hypercomplete. At the moment, however, I don't see whether Booleanness implies any of these other conditions.
Thus my question is: can a Boolean $(\infty,1)$-topos fail to be hypercomplete?