The following is a counterexample for $\mathcal{F}$ a presheaf of abelian groups (or sets, if you like).
Let $X=\lbrace a,b,c,d\rbrace$ with nontrival opens given by $\lbrace a \rbrace,\lbrace b \rbrace,U=\lbrace a,b,c \rbrace,V=\lbrace a,b,d \rbrace, U\cap V$.
Define the presheaf $\mathcal{F}$ by
$\mathcal{F}(\lbrace a \rbrace)=\mathcal{F}(\lbrace b \rbrace)=\mathbb{Z}/2\mathbb{Z}$,
$\mathcal{F}(U)=\mathcal{F}(V)=\mathcal{F}(U\cap V)=\mathcal{F}(X)=\mathbb{Z}$,
with the obvious restriction maps.
Then $\mathcal{F}^+(X)=\lbrace (x,y)\in\mathbb{Z}\oplus\mathbb{Z}| x\equiv y\text{ (mod 2)}\rbrace$, since the germs at $a$ and $b$ are determined by those at $c$ and $d$, and the only restrictions on $c$ and $d$ are that they give the same germs at $a$ and $b$.
Consider $(0,2)\in\mathcal{F}^+(X)$. The germs at $c$ and $d$ cannot come from a common section of $\mathcal{F}(X)$. Any system of sections which does not include $X$ in the cover must include both $U$ and $V$, having sections 0 and 2, respectively. Of course these do not agree when restricted to $U\cap V$. QED
This construction relies crucially on the fact that the presheaf is not separated (i.e. gluing is not unique). If the presheaf $\it{is}$ separated, the condition described in the question is clearly satisfied.
This construction was shown to me by Paul Balmer.
Here is a sketch of why I think the condition that $Y$ is a local homeomorphism over $X$ should be sufficient for the counit to be a homeomorphism. I haven't worked out the converse yet.
For a presheaf $F \in Set^{\mathcal{O}(X)^{op}}$, the formula for the left Kan extension should be $$L(F) = \mathrm{colim}_{y(U) \rightarrow F} U$$ where $y : \mathcal{O}(X) \rightarrow Set^{\mathcal{O}(X)^{op}}$ is the Yoneda embedding. By Yoneda's lemma, the indexing category for the colimit is exactly the category of elements of $F$ which I will write as $\int F$.
Now, consider the case where $F = \Gamma_Y$ for some space $p: Y \rightarrow X$ and assume that $p$ is a local homeomorphism. We have $\Gamma_Y(U) = \{\sigma : U \rightarrow Y | p \circ \sigma = \mathrm{id}_U \}$. So the objects in the category $\int \Gamma_Y$ are exactly the sections over the various open sets of $X$, and the morphisms are given by restriction of sections. I'll write $d(\sigma)$ for the domain of a given section.
Our Kan extension formula becomes $$L(\Gamma_Y) = \mathrm{colim}_{\sigma \in \int \Gamma_Y} d(\sigma)$$ From here it's easy to see what the counit is: since our object is given by a colimit, it suffices to construct a map $d(\sigma) \rightarrow Y$ for each $\sigma \in \int \Gamma_Y$. But clearly $\sigma$ itself qualifies.
Now choose an open covering $\{V_\alpha\}$ of $Y$ such that $p$ restricts to a homeomorphism on each $V_\alpha$. We then have a collection $\{\sigma_\alpha : p(V_\alpha) \rightarrow V_\alpha\}$ of sections by choosing the inverse to each restriction. My claim would be that this collection is cofinal (or final? I can never remember which) in the category $\int \Gamma_Y$ so that we can restrict our colimit to just this subcategory. Notice that in this case, the components of the counit map above are homeomorphisms.
Moreover, this subcategory should also be cofinal in $\mathcal{O}(Y)$ by associating $\sigma_\alpha$ with the open set $V_\alpha$. Then the fact that the counit is a homeomorphism should be the statement that a topological space is the colimit of any of its open coverings.
Is this along the lines of what you were thinking?
Best Answer
A presheaf $F$ with values in $C$ is a called a sheaf if, for every object $X$ and every covering sieve $R$ of $X$, the natural maps
$F(X) \rightarrow F(Y)$
for each Y in R induce an isomorphism
$F(X) \xrightarrow{\sim} \varprojlim_{Y \in R} F(Y)$
This definition makes sense without any assumptions on $C$.
The sheafification construction makes use of filtered colimits and essentially arbitrary limits (if you are interested in sheaves on a particular topology then you might be able to restrict the class of limits that need to be considered). It is defined by iterating the construction
$F^+(X) = \varinjlim_{R} \varprojlim_{Y \in R} F(Y)$
where the $\varinjlim$ is taken over covering sieves of $X$. If $F$ is set-valued, the associated sheaf of $F$ is $F^{++}$.
I don't know what conditions on $C$ are necessary to make the sheafification of a presheaf in $C$ a sheaf, but I wouldn't expect the construction to behave very well unless $C$ is a fairly special category.
(Categories of algebraic structures on sets defined by finite inverse limits would qualify as "special", essentially because filtered colimits commute with finite inverse limits.)