I think, it does.
By change of coordinates, you may suppose that $Z$ contains the origin and the tangent vector space is the hyperplane $\{x_n=0\}$. Then, by implicit function theorem, for small enough $x=(x_1,\ldots,x_{n-1})\in \mathbb{R}^{n-1}$ the polynomial $q_x(t):=Q(x,t)$ has a real root, and $p_x(t):=P(x,t)$ has the same root. So, the resultant of $p_x$ and $q_x$ is 0 for small enough $x$. Since this resultant is a polynomial in $x$, it is identical 0. Consider the polynomials $p(t)=:P(x_1,\ldots,x_{n-1},t)$, $q(t):=Q(x_1,\ldots,x_{n-1},t)$ over the field $\mathbb{R}[x_1,\ldots,x_{n-1}]$. Their discriminant is 0. I claim that $q$ is irreducible. Indeed, if $q(t)=a(t)b(t)$ with non-constant $a$, $b$, then multiplying by the common denominator we get a formula $Q(x_1,\ldots,x_{n-1},t)D(x_1,\ldots,x_{n-1})=A(x_1,\ldots,x_{n-1},t)B(x_1,\ldots,x_{n-1},t)$ for real polynomials $A,B$. Thus, since $Q$ is irreducible, one of $A$, $B$ must be divisible by $Q$ (in $\mathbb{R}[x_1,\ldots,x_n-1,t]$), but this is not the case as each of $A$, $B$ has smaller degree then $Q$ w.r.t. variable $t$.
So, $q(t)$ is irreducible, but the discriminant of $p(t)$ and $q(t)$ is 0. This yields that $q$ divides $p$ that after multiplication by the denominator yields that $P(x_1,\ldots,x_{n-1},t)H(x_1,\ldots,x_{n-1})$ is divisible by $Q$. But if $Q$ divides $H$, then $Z$ does not have a tangent space of the form $\{x_n=0\}$ (quite the opposite), so $Q$ divides $P$.