Here is some idea, it is not very precise. Let $F$ be the free group on two generators and let $F_p$ be its pro-$p$ completion. Let $w$ be an infinite word in $F_p$, i.e., an element of $F_p$ which is not in $F$. Let $W$ be the closed verbal subgroup of $F_p$ generated by $w$. I think that it is possible to choose $w$ so that $W \cap F$ is trivial. For example, the first congruence subgroup of $SL_2(\mathbb{Z}_p)$ ($P>2)$ satisfies a pro-$p$ identity due to Zubkov, but I think it has a dense free group.

Let $D_n$ be the $n$-dimension subgroups of $F_p$. Then if I recall correctly there is a canonical way to write $w=w_nu_n$, where $u_n \in D_n$. Take your variety to be all the groups that satisfy $w_n$ for some $n$. I would guess they will satisfy your requirement.

I am not sure this idea will work, but if you like it and need more reference, then please contact me.

Here is some idea, it is not very precise. Let $F$ be the free group on two generators and let $F_p$ be its pro-$p$ completion. Let $w$ be an infinite word in $F_p$, i.e., an element of $F_p$ which is not in $F$. Let $W$ be the closed verbal subgroup of $F_p$ generated by $w$. I think that it is possible to choose $w$ so that $W \cap F$ is trivial. For example, the first congruence subgroup of $SL_2(\mathbb{Z}_p)$ ($p>2)$ satisfies a pro-$p$ identity due to Zubkov, but I think it has a dense free group.

Let $D_n$ be the $n$-dimension subgroups of $F_p$. Then if I recall correctly there is a canonical way to write $w=w_nu_n$, where $u_n \in D_n$. Take your variety to be all the groups that satisfy $w_n$ for some $n$. I would guess they will satisfy your requirement.

I am not sure this idea will work, but if you like it and need more reference, then please contact me.