$\DeclareMathOperator{\scrV}{\mathcal{V}}$Yes. As a test case consider the case that $n = 2$ and $\scrV$ is a countable collection of points. Then for any $d \geq 2$, the space $P_d$ of homogeneous polynomials of degree $d$ vanishing at $p$ is a positive dimensional vector space, and there is polynomial $F \in P_d$ which does not vanish at any other point of $\scrV$, since a vector space over an uncountable field can not be the union of countably many proper subspaces. 

The general case follows in the same way. We can of course assume each $V$ is irreducible (which you probably mean anyway). For each $V \in \scrV$, choose a point $p_{1, V} \in V \setminus \{p\}$. By the same argument as above, there is a homogeneous polynomial $F_1$ which vanishes at $p$, but does not vanish on any $p_{1, V}$. Let $\scrV_2$ be the countable collection of irreducible components of $V(F_1) \cap V$ for all $V \in \scrV$. Choose $p_{2,V} \in V \setminus \{p\}$ for all $V \in \scrV_2$, and continue as above. 

After step $m+1$, where $m$ is the largest dimension of all varieties in $\scrV$, you will have that $Z := V(F_1) \cap \cdots \cap V(F_{m+1})$ satisfies the property you seek. Note that $\dim(Z) \geq n - m - 1 \geq 1$.