Can natural section/retraction be checked pointwise?

No, you cannot check the property of being a section/retraction pointwise. Take $C = {\cdot \to \cdot}$ so that the category of functors from $C$ to $D$ is the arrow category of $D$. In the arrow category, the property of an object being an isomorphism (as a morphism of $D$) is closed under retracts (exercise). Now let $f : A \to B$ be a morphism of $D$ which admits a retraction but is not an isomorphism. Then there is an obvious morphism in the arrow category of $D$ from the object $(f : A \to B)$ to the object $(1 : B \to B)$ and it is pointwise the inclusion of a retract. But it cannot be the inclusion of a retract in the arrow category, as then the original map $f$ would be an isomorphism.

This example shows that the obvious sufficient condition on $D$ (namely, every retraction is already an isomorphism) is required. But perhaps you prefer some concrete examples for intuition. Then consider, in the category of $G$-sets, the map $G \to *$, which has no section; or in the category of simplicial sets, the map $\partial \Delta^1 \to \Delta^1$, which is not the inclusion of a retraction.

Here's an example I find easier to think about than the examples given so far. Let $G$ be a group and $k$ a field, let $C = BG$ be the category with one object with automorphisms $G$, and let $D = \text{Vect}(k)$ be the category of $k$-vector spaces. Then the functor category $[C, D]$ is the category of linear representations of $G$ over $k$.

Your question in this special case, for retracts, is equivalent to asking whether every subrepresentation $V \subseteq W$ is a direct summand as a representation (it is always a direct summand as a vector space). This is true iff the group algebra $k[G]$ is semisimple, which is true iff $G$ is finite and the characteristic of $k$ does not divide $|G|$. So for an explicit counterexample we can either take $G = \mathbb{Z}$ and consider a nontrivial Jordan block, or take $G = C_p, k = \mathbb{F}_p$ and, well, again consider a nontrivial Jordan block.

[Note: this post does not answer the question, which was whether it is possible to give $\epsilon$ which has a right inverse, but every right inverse is non-natural.]

To provide a counter-example, let us consider the category of sets and the constant functors $F(X) = 2 = \{0,1\}$, $G(X) = 1 = \{0\}$. There is precisely one transformation $\epsilon : F \Rightarrow G$, namely $\epsilon_X(x) = 0$, and it is natural.

Every transformation $\eta : G \Rightarrow F$ is a right inverse of $\epsilon$, but not every such $\eta$ is natural. For instance, take $$\eta_X(0) = \begin{cases} 0 & \text{if $X = \emptyset$} \\ 1 & \text{otherwise}. \end{cases} $$ Then naturality of $\eta$ fails for the map $f : \emptyset \to 1$.