Is sheafification always an inclusion?

No, $\mathcal F \to \mathcal F^+$ is not injective in general.

Consider the sheaf $\mathcal C$ of continuous functions on $\mathbb R$, its subpresheaf $\mathcal C_b\subset \mathcal C$ of bounded continuous functions and the quotient presheaf $\mathcal F$, characterized by $\mathcal F(U)= \mathcal C (U)/\mathcal C_b(U)$.
For every open subset $U\subset \mathbb R$, we have $\mathcal F(U)\neq 0$ but the associated sheaf is $\mathcal F^+=0$, so that
the morphism $\mathcal F \to \mathcal F^+=0$ is definitely not injective.

Injectivity of $\mathcal F \to \mathcal F^+$ is equivalent to requesting that whenever you have compatible gluing data $s_i\in \mathcal F(U_i)$ on an open covering $(U_i)$ of an open $U$ of your space, they can glue to at most one $s\in \mathcal F(U)$ : one half of the conditions for a presheaf to be a sheaf must be satisfied (the other half is to require that $s$ always exist)

The answer to the question is NO since the global information about a presheaf can not be determined by its local information. In fact, this is the essential point of the definition of a sheaf. In above, Georges Elencwajg has given a nice counter-example which is quite natural. We can also contruct a counter-example directly as follows. Take a topological space $X$ which is not empty, define a presheaf $\mathcal{F}$ of abelian groups in the following way: for any proper open subset $U$ define $\mathcal{F}(U)$ to be zero while define $\mathcal{F}(X)$ to be any nontrival abelian group, say $\mathbb{Z}$. Then the sheafication $\mathcal{F}^+$ is zero. Consider the global sections we will find it is not an inclusion.

Cokernels do not have this property. The example to look at is the exponential map from the sheaf of holomorphic functions on $\mathbf C \setminus \{0\}$ to the sheaf of non vanishing holomorphic functions on that same domain. This is not surjective on global sections, since for example the identity map does not have a logarithm, but it is surjective as a morphism of sheaves, so the sheaf cokernel is trivial.

What distinguishes the kernel and image sheaves, I think, is that they satisfy the identity axiom [I think this is Hartshorne's (5)] because they sit inside of sheaves.