Yes, in fact Grothendieck fibration between groupoids are enough.

Let $p:Y \to X$ be any surjection.

We construct the following groupoid $G$. Its set of objects is $X \coprod Y$. Its morphisms corresponds to the equivalence relation such that two elements of $y$ are equivalent if they have the same image by $p$ and an element of $y$ is equivalent to its image by $p$ in $X$ (and no distinct elements of $X$ are equivalents).


We then consider $I$ the walking isomorphisms, i.e. the anti-discrete groupoid on two objects, and the functor $G \to I$ that sends element of $Y$ on one object and element of $X$ on the other.

I claim that:

1) it is a fibration. (easy to check)
2) a Cleavage produces a section of $p$: for each $x \in X$ a cartesian lift of the non-identity arrow $e \to p(x)$ corresponds to the choice of a $y \in Y$ such that $p(y)=x$.