Suppose $\Bbbk$ is a finite field of size $q$. Let $V_{i,n}= \mathfrak{p}_i \cap R_n$, $d_n= \dim_{\Bbbk} R_n$ and $d_{i,n}=\dim_{\Bbbk}V_{i,n}$. It is enough to show that for $n$ big enough:

$$|V_n| > \sum_1^k |V_{i,n}| $$

Note that for each $i$, $d_n - d_{i,n}$ gives the Hilbert function of $R/\mathfrak{p}_i$, so it eventually becomes a polynomial in $n$ of dimension $=\dim R/\mathfrak{p}_i \geq 1$. So for $n\gg 0$, $q^{d_n-d_{i,n}}>k$ for each $i$. Thus

$|V_n|/|V_{i,n}|= q^{d_n-d_{i,n}}> k$

and we are done.

EDIT: the argument below only works if each $R/\mathfrak{p}_i$ has dimension at least $2$, as pointed out by Angelo.

Suppose $\Bbbk$ is a finite field of size $q$. Let $V_{i,n}= \mathfrak{p}_i \cap R_n$, $d_n= \dim_{\Bbbk} R_n$ and $d_{i,n}=\dim_{\Bbbk}V_{i,n}$. It is enough to show that for $n$ big enough:

$$|V_n| > \sum_1^k |V_{i,n}| $$

Note that for each $i$, $d_n - d_{i,n}$ gives the Hilbert function of $R/\mathfrak{p}_i$, so it eventually becomes a polynomial in $n$ of dimension $={\dim R/\mathfrak{p}_i}-1 \geq 1$ (EDIT: I originally wrote $\dim R/\mathfrak{p}_i$). So for $n\gg 0$, $q^{d_n-d_{i,n}}>k$ for each $i$. Thus

$|V_n|/|V_{i,n}|= q^{d_n-d_{i,n}}> k$

and we are done.