The Hilbert scheme is pretty horribly behaved, and positive results of this nature are quite rare.

You probably mean to ask if there exists $C\in L$ such that $C$ is both reduced and irreducible. If you don't demand reducedness, the answer is yes. For suppose $H$ is a component of a Hilbert scheme of curves in $\mathbb{P}^n$, and let $C\in H$ be a general member of $H$. Let $M_t\in PGL(n+1)$ be a family of linear transformations which limits to the projection from a general codimension 2 plane in $\mathbb{P}^n$ as $t\to 0$ (so that it projects onto a line). Then $M_t(C)$ gives a curve in $H$, but the limiting curve in $H$ is supported entirely on a line, hence is irreducible.

Whenever you have a component with a reduced and irreducible member, it is in fact true that the general member is reduced and irreducible, as being integral is open in flat families.

In general, the Hilbert scheme is always connected (a theorem of Hartshorne); however, irreducible components typically meet at highly non-reduced members instead of the "nice" objects we're actually trying to parameterize.

It is not reasonable to expect that every component of a Hilbert scheme of curves has an irreducible member. For instance, the Hilbert scheme parameterizing twisted cubic curves also has a separate component parameterizing unions of plane cubic curves and an isolated point. You may object this latter object is not a "curve," but part of the problem with Hilbert schemes is that the Hilbert scheme of curves doesn't even only see things of pure dimension 1.

Similarly, the Hilbert scheme corresponding to the Hilbert polynomial of two skew lines does not have any reduced and irreducible curves in it.

To get more satisfying counterexamples, one only needs to increase the degree and genus--things get really nasty really fast.