I would guess not. Here is some evidence towards this guess.

Let $S$ be a $p$-adically complete ring and let $M$ be $p$-adically complete $S$-module which is free over $S$. Then I claim that $M$ must be finitely generated over $S$.

To see this, suppose that $X$ is an infinite free generating set for $M$ over $S$: $M = \bigoplus_{x \in X} Sx$. Choose some infinite sequence $x_0, x_1, \cdots$ of elements in $X$. Then the partial sums $\sum_{i=0}^n p^i x_i$ form a Cauchy sequence in the $p$-adic topology of $M$ but do not converge to an element of $M$, a contradiction.

So $\mathbb{Z}_p \langle x,y \rangle$ cannot be free over $\mathbb{Z}_p \langle x \rangle$.

I would guess not. Here is some evidence towards this guess.

Let $S$ be a $p$-adically complete ring of characteristic $0$ and let $M$ be $p$-adically complete $S$-module which is free over $S$. Then I claim that $M$ must be finitely generated over $S$.

To see this, suppose that $X$ is an infinite free generating set for $M$ over $S$: $M = \bigoplus_{x \in X} Sx$. Choose some infinite sequence $x_0, x_1, \cdots$ of elements in $X$. Then the partial sums $\sum_{i=0}^n p^i x_i$ form a Cauchy sequence in the $p$-adic topology of $M$ but do not converge to an element of $M$, a contradiction.

So $\mathbb{Z}_p \langle x,y \rangle$ cannot be free over $\mathbb{Z}_p \langle x \rangle$.