Take $p=2x_1x_2$, and $U=U_r$ a small disk of radius $r$ in $S^{n-1}$ in centered at $(1,0,\dotsc, 0)$. $\newcommand{\pa}{\partial}$ Then $\nabla p(x)=2x_2\pa_{x_1}+2x_1\pa_{x_2}$ so that

$$ \sup_{x\in U_r}|\nabla p(x)|\geq 2.$$

On the other hand,

$$\sup_{x\in U_r}|p(x)|= O(r). $$ Thus

$$ \lim_{r\to 0}\frac{\sup_{x\in U_r}|\nabla p(x)|}{\sup_{x\in U_r}|p(x)|}=\infty. $$

Thus the constant $C(p)$ has to depend on the region $U$ as well. On the other hand, you can find a constant that is *independent* of $p$, but dependent on $U$ and on the degree of $p$.

Denote by $H_d$ the space of degree $d$ homogeneous polynomials in $n$ variables. For $p\in H_d$ define

$$\Vert p\Vert_{U,0}:=\sup_{x\in U} |p(x)|,$$

$$\Vert p\Vert_{U,1}:=\sup_{x\in U}|\nabla p(x)|. $$

Because ${\rm vol}\,(U)>0$ we have

$$ p\in H_d,\;\;\Vert p\Vert_{U,0}=0 \Longleftrightarrow p=0. $$

Indeed, the set $\{p=0\}\cap S^{n-1}$ is semialgebraic and, if it contains the set $U$ of nonzero volume (in $S^{n-1}$), it also contains an open subset of $S^{n-1}$. By unique continuation, $p$ must be identically zero on$ S^{n-1}$.

Similarly, if $\Vert p\Vert_{U,1}=0$ we deduce that $\nabla p=0$ on $U$. Arguing as before we deduce that $\nabla p=0$ on $S^{n-1}$ and, in particular $p$ is constant on $S^{n-1}$. Since $p$ is a homogeneous polynomial of degree $d$, we deduce that it must have the form $p(x)= C|x|^{d/2}$. The gradient of this function is not zero if $d>0$. We thus have

$$ d>0\;\; p\in H_d,\;\;\Vert p\Vert_{U,1}=0\Longleftrightarrow p=0. $$

Thus, when $d>0$, the functions $\Vert p\Vert_{U,0}$ and $\Vert p\Vert_{U,1}$ are norms on the finite dimensional vector space $H_d$ and, as such, they are equivalent.