Set $p$ prime.
Take $f_1, f_2 \in \mathbb{Z}$ with $f_1 f_2$ square free and $p \nmid f_1, f_2$ and $\alpha \in \mathbb{Q}_p$ such that $|\alpha|_p = 1$ (i.e. $\alpha$ is a $p$-adic unit).
Show that $|f_1 + \alpha^{2}f_2|_{p} < 1 \implies \exists r \in \mathbb{Z}. \; f_1 + r^{2}f_2 \equiv 0 \; (p)$.
This is clear in the case $\alpha \in \mathbb{Q} \cap \mathbb{Q}_p$, as then:
$\quad \exists u,v \in \mathbb{Z}. \; \alpha = \frac{u}{v} \quad \quad p \nmid u, v$
so:
$\quad |f_1 + \alpha^{2}f_2|_{p} = \left|f_1 + \frac{u^2}{v^2}f_2\right|_{p} = \left|\frac{v^{2}f_1 + u^{2}f_2}{v^2}\right|_{p} = |v^{2}f_1 + u^{2}f_2|_{p} < 1$
then $p \; | \; v^{2}f_1 + u^{2}f_2$, so:
$\quad v^{2}f_1 + u^{2}f_2 \equiv 0 \; (p)$
But as $p \nmid v \implies gcd(p,v) = 1 \implies \exists w \in \mathbb{Z}. \; vw \equiv 1 \; (p)$, so:
$\quad f_1 + (wu)^{2}f_2 \equiv 0 \; (p)$
So $\alpha = wu$ works.
However, I am not clear on the case $\alpha \in \mathbb{Q}_{p} \backslash \mathbb{Q}$.
edit: removed irrelevant piece of information.
Your proof doesn't use the fact that $\alpha \in \mathbf{Q}$; all you actually use is that $u,v \in \mathbf{Z}_p$.
(and so, your proof can be simplified, since $p \not\mid v$ means that $\alpha \in \mathbf{Z}_p$, so you can assume $v=1$)
Modular arithmetic can be defined over $\mathbf{Z}_p$ in the usual way:
$$ x\equiv y \pmod{z} \Leftrightarrow z \mid x-y \Leftrightarrow \frac{x-y}{z} \in \mathbf{Z}_p $$
For the $p$-adic numbers specifically, this relation can be written in terms of the $p$-adic norm as well: $\|x - y\|_p \leq \|z\|_p$.
The only relevant quality of $z$ is how many powers of $p$ divide it. This even looks like modular arithmetic over the integers: e.g. there are only $p^n$ residue classes of $p$-adic integers modulo $p^n$.
You can even define modular arithmetic over $\mathbf{Q}_p$ with the same caveats you would use when defining it over $\mathbf{Q}$: again,
$$x \equiv y \pmod{z} \Leftrightarrow \frac{x-y}{z} \in \mathbf{Z}_p $$
however, you have to use the full form of multiplication: if $d \neq 0$, then
$$ a \equiv b \pmod{c} \Leftrightarrow ad \equiv bd \pmod{cd} $$
Over $\mathbf{Z}$ (or $\mathbf{Z}_p$) you can ignore the $d$ in the modulus, but you can't do that if $d$ is not integral. In particular, you can't multiply two congruences, because
$$ a \equiv c \pmod{m} \qquad b \equiv d \pmod{m}$$
would imply
$$ ab \equiv cb \pmod{bm} \qquad cb \equiv cd \pmod{cm} $$
and you can't chain the equivalences together with transitivity. Ultimately, you get
$$ ab \equiv cd \pmod{m \, \gcd(b,c)} $$