A generic curve of large genus, say at least $24$, will not be a fibre of a Lefschetz pencil. 

The reason is that by the theorems of Harris-Mumford and Eisenbud-Harris, the moduli space of curves of large genus is of general type, so there can be no rational curve passing through a generic point. To use this one needs to know that all the smooth fibres of the pencil are not isomorphic. Since the local monodromy around a singular fibre is infinite by the Picard-Lefschetz formula, it suffices to show that there must be at least one singular fibre. But if all fibres are smooth, then $\tilde{Y}$ (the total space of the Lefschetz pencil) must be isomorphic to $C \times \mathbb{P}^1$. This cannot happen since  $C \times \mathbb{P}^1$ is not a blow up of any other surface.

A similar argument should work for other varieties whose moduli spaces are of general type.