First of all, is it true?
Every presheaf $F:\mathcal{C}^{\text{op}}\to\mathbf{Set}$ can be written as a colimit of representable functors. It can be expressed as a coend, $$\int^{A\in\mathcal{C}} F(A)\times \hom_{\mathcal{C}}(-,A).$$ Can we write a copresheaf $G:\mathcal{D}\to\mathbf{Set}$, similarly, as an end?
The answer to the first question is yes.
You can easily observe that for each object $X \in \mathcal C$ we have the mappings
$$i_1^X \colon \int^{A \in \mathcal C} F(A) \times \hom_{\mathcal C}(X,A) \longrightarrow F(X)$$ $$[(x,\alpha)] \stackrel{i_1^X}{\mapsto} F(\alpha)(x)$$ $$i_2^X \colon F(X) \longrightarrow \int^{A \in \mathcal C} F(A) \times \hom_{\mathcal C}(X,A)$$ $$x \stackrel{i_2^X}{\mapsto} [(x,1_X)]$$ (where by $[(x,\alpha)]$, with $x \in F(A)$ and $\alpha \in \hom_{\mathcal C}(X,A)$, I denote the equivalence class of the pair $(x,\alpha) \in F(A) \times \hom_{\mathcal C}(X,A)$ under the equivalence relation defining the coend).
It is a matter of computation to prove that these mappings are well defined and natural in $X$.
Also a matter computation, but more interesting, these maps are inverse to each other: $$i_2(i_1([x,\alpha]))=i_2(F(\alpha)(x))=[(F(\alpha)x,1_X)]=[x,\alpha]$$ (this equality follows by the properties of coends) $$i_1(i_2(x))=i_1([x,1_X])=F(1_x)(x)=1_{F(X)}(x)=x\ .$$
So the $i_1$ and $i_2$ give natural isomorphisms between the functors $F$ and $\int^{A \in \mathcal C}F(A) \times \hom_{\mathcal C}(-,A)$.
The answer to the second question
is yes too.is no, to be more exact what it holds is that $G \cong \int^{A \in \mathcal D}G(A)\times \hom_{\mathcal D}(A,-)$, where the functor on the right is still a coend.You can regard a functor $G \colon \mathcal D \to \mathbf{Set}$ as a presheaf $G \colon \mathcal {D^{\text{op}}}^\text{op} \to \mathbf {Set}$, then you have that $$G \cong \int^{A \in \mathcal D^\text{op}} G(A) \times \hom_{\mathcal D^\text{op}}(-,A) \cong \int^{A \in \mathcal D}G(A) \times \hom_{\mathcal D}(A,-)$$ where the isomorphisms are natural.