The ring of integers of a non-archimedean local field $K$ of characteristic $0$ (i.e. an extension of $\Bbb Q_p$ for a unique prime number $p$) is the integral closure of $\Bbb Z_p$ in $K$.
This is the same situation as for number fields: when $L$ is a number field, the ring of integers $O_L$ is the integral closure of $\Bbb Z$ in $L$.
We can further notice that $\Bbb Z$ is dense in $\Bbb Z_p$ (and even in $\hat{\Bbb Z}$). The $p$-adic unit ball $\Bbb Q \cap \Bbb Z_p$ in $\Bbb Q$ is the localization $\Bbb Z_{(p)}$, which is a DVR (hence PID, Dedekind, UFD, integrally closed, ...).
But the ring of integers $\Bbb Z$ is not a "unit ball" ; it is rather an intersection of the unit balls $\Bbb Z_{(p)}$ with $p$ varying among the primes:
$$\Bbb Z = \bigcap_p \Bbb Z_{(p)}.$$
This can be generalized to the ring of integers of any number field, as shown here.
In general, if $(K,v)$ is a valued field, we can define the valuation ring as $O_K = \{x \in K \mid v(x) \geq 0\}$. In particular,
$$\Bbb Z_p = \{ x \in \Bbb Q_p \mid v_p(x) \geq 0 \},$$
and notice that $v_p(x) \geq 0 \iff |x|_p \leq 1$.
The condition $v_p(x) \geq 0$ means that $x$ has a "denominator" not divisible by $p$ (this makes sense at least if $x \in \Bbb Z$). Any other prime $q \neq p$ is invertible in $\Bbb Z_p$, so intuitively the condition $v_p(x) \geq 0$ should mean that $x$ has no denominator, i.e. should be considered as a ($p$-adic) integer.
It might be useful to mention that $\Bbb Q_p = \Bbb Z_p[1/p]$ (as algebras), and any $z \in \Bbb Q_p^{\times}$ can be written uniquely as
$z = p^n u$ with $u \in \Bbb Z_p$ and $n \in \Bbb Z$. The condition $|x|_p \leq 1$ (equivalently $v_p(x) \geq 0$) just means that $n \geq 0$, whereas $n = -1$ for instance gives $z = u / p$, which is not a $p$-adic integer, having $p$ as "denominator".
