= Solution
Write $p(z)=c\prod_{j=1}^{2m}(z-a_j)$, including root multiplicities. For $|z|>R$ the convergent logarithm series for every factor $1-a_j/z$ defines
$$
L_j(z)=-\sum_{n\ge1}\frac{a_j^n}{n z^n},\qquad e^{L_j}=1-a_j/z.
$$
Choosing either square root of the nonzero leading coefficient gives the <holomorphic square root outside all polynomial roots>
$$
\boxed{h(z)=\sqrt c\,z^m\exp\left(\tfrac12\sum_j L_j(z)\right).}
$$
Its square is $p$. The even degree makes $z^m$ single-valued, avoiding the winding obstruction that an odd degree would create on the exterior annulus.
For the branch behaving as $z^2$ at infinity, the <Laurent series> is
$$
\sqrt{z^4-z}=z^2(1-z^{-3})^{1/2}
=z^2-\tfrac12z^{-1}-\tfrac18z^{-4}-\cdots.
$$
It converges uniformly on the circle of radius two. Only the coefficient of $z^{-1}$ contributes to its anticlockwise integral, so
$$
\boxed{\int_C\sqrt{z^4-z}\,dz=-\pi i}
$$
for this branch, and $+\pi i$ for the other. These are the two allowed answers.
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