The following interesting question came up in a discussion I was having with Alex Wright.
Suppose given a branched cover C -> P^1 with four branch points. It's not hard to see that the field of definition of C has transcendence degree at most 1 over $\mathbf{\bar{Q}}$.
Which leads one to ask:
Give an example of a field K of transcendence degree 1 over $\mathbf{\bar{Q}}$, and a geometrically connected curve C/K, such that C does not admit a branched covering to P^1 with four branch points.
Update: Ben Weiland suggests that taking C to be the universal curve over the function field of a a compact curve in $M_g$ might give such an example. Indeed, one might ask for lower bounds on the number of singular fibers (or, thinking of a curve in $\bar{M}_g$, on the intersection with various boundary components.)
Re-update: Both Ben and Jason present examples of compact families of 4-branched covers. So I think the question still stands open as written. Based on the discussion in the answers and comments, I would think that a promising candidate would be provided by a compact Shimura curve parametrizing genus-2 curves whose Jacobians have multiplication by an indefinite quaternion algebra. Can this be a component of a Hurwitz curve in M_2? (Warning: as Dan Peterson points out these are not actually compact in M_2, only in A_2.)
Some remarks:
$\bullet$ If you replace "four branch points" with "three branch points" and "transcendence degree 1" with "transcendence degree 0," the nonexistence of such an example is Belyi's theorem.
$\bullet$ There is no obstruction coming from the choice of K; a theorem of Diaz, Donagi, and Harbater guarantees that for any field K of transcendence degree 1 over $\mathbf{\bar{Q}}$, there exists a geometrically connected curve C/K which admits a 4-branched cover to P^1.
$\bullet$ There is a somewhat subtle issue which doesn't arise in the Belyi case: a 4-branched cover C/K->P^1/K yields a map from Spec K to $\mathcal{M}_{{0,4}}$
("forget the cover, remember the branch points") which, after passage to the generic point gives you a choice of inclusion of the function field $\mathbf{\bar{Q}}(\mathcal{M}_{0,4})$ into K.
One can either consider K as an abstract field, or as a fixed extension of the rational function field $\mathbf{\bar{Q}}(\mathcal{M}_{0,4})$, which places a stronger condition on C. E.G. if you ask that this inclusion be an isomorphism, you are requiring that C be a 4-branched cover which is "determined uniquely by its branch points," like y^2 = x(x-1)(x-t).
Best Answer
Edit: this doesn't work.
Original: I think that a compact curve in $M_g$ provides an example. In such a family, the length of the shortest geodesic is bounded. Whereas, in a family produced by ramifying over four points in $P^1$, as two ramification points collide, there is a loop whose image downstairs stays near that pair of points, looping about them several times, and thus has arbitrarily small length. There's probably a variant of this argument with Mumford-Tate groups.
Added:
Jason objects that branch points downstairs can collide without ramification points upstairs colliding. This is true, but easy to patch. There are four branch points and some pairs of them might collide without the ramification points colliding, but if all pairs collide without the ramification points colliding, then they must permute different sets of sheets and the curve is not (geometrically) connected.
Jordan objects that even if the ramification points collide, the curve may remain smooth. In particular, if the curve is a $d$-fold cover of $P^1$ and all the monodromy is a power of a fixed $d$-cycle, if two points labeled by $a$ and $b$ so that all three of $a$, $b$, and $a+b$ are relatively prime to $d$, then the family is smooth. If this is true for all pairs of collisions, this gives a complete family of curves that map to $P^1$ with only four branch points, contradicting my claim. In particular, $d=5$ and $1,1,1,2$ is a complete family of genus $4$ curves.
Such examples don't exist for smaller $d$. Thus a complete curve in $M_3$ (which exist, right?) is a good candidate for not having a map to $P^1$ with $4$ branch points. Collisions of branch points with other forms of monodromy are harder for me to understand, but they look more likely to result in degeneration.