$\def\bQ{\mathbb{Q}}\def\bF{\mathbb{F}}\def\bZ{\mathbb{Z}}$This is false as stated because of the following important difference between $K$ and $\mathbb{C}$: the former is not algebraically closed. For example. consider $K=\bQ_p(p^{1/k})$ with $k>1$ and the polynomial $$f(z)=z\prod\limits_{a\in\bF_p^{\times}}(z-[a])^2$$ with $r_1=1/p,r_2=1$(here $[x]$ denotes the Teichmuller representative of an element $x$ of the residue field of a complete non-archimedean field). Then for any $z\in K$ with $|z|$ equal to $1$ we have $|z-[a]|\leq p^{-1/k}$ for exactly one $a$ because the residue field of $\mathcal{O}_K$ is equal to $\bF_p$, so $|f(z)|\leq p^{-2/k}$. For $z$ with $|z|=p^{-1}$ we just get $|f(z)|=p^{-1}$. However, taking $z=p^{1/k}$ gives $|f(z)|=p^{-1/k}$ which is larger than any value of $f$ on the boundary.
The statement becomes true if we assume that the residue field of $\mathcal{O}_K$ is infinite(the counterexample above depends crucially on the finiteness of the residue field). The formation of the Newton polygon is a very convenient way to visualize the behavior of roots of a $p$-adic analytic function and the desired inequality in the case of an algebraically closed field $K$ follows quickly from the fact that the slopes of the Newton polygon of a Laurent series correspond to the valuations of the roots.
However, It might be instructive to unpack the proof of this theorem to get a direct argument for our boundary inequality: let $f(z)=\sum\limits_{n\in \bZ}a_nz^n$ be a Laurent series converging for $z$ satisfying $r_1\leq |z|\leq r_2$. Assume that $r_1,r_2$ are in the image of the norm map on $K$. We will prove that for every such $z$ there is $u$ with $|u|$ equal to $r_1$ or $r_2$ such that $|f(z)|\leq |f(u)|$.
Lemma. If the residue field of $\mathcal{O}_K$ is infinite, then for every $z\in K$ there exists an element $z'\in K$ with $|z|=|z'|$
and a number $n\in\bZ$ such that $|a_n(z')^n|$ is larger or equal to
any $|a_m(z')^m|$ with $m\neq n$ and $|f(z)|=|a_n(z')^n|$.
Proof.
Since the values $|a_nz^n|$ tend to zero as $n$ tends to $\pm\infty$,
there exists a finite set of indices $i_1<\dots< i_k$ such that
$|a_{i_1}z^{i_1}|=\dots=|a_{i_k}z^{i_k}|$ and $|a_mz^m|$ is less that
this common value for any $m\notin\{i_1,\dots, i_k\}$. We want to find
$z'$ such that the norm of the sum of these $k$ summands is precisely
equal to the norm of each separate summand. To arrange that, pick
$\lambda\in \mathcal{O}_K/\mathfrak{m}_K$ such that
$1+\lambda^{i_2-i_1}\rho\left(\frac{a_{i_2}z^{i_2}}{a_{i_1}z^{i_1}}\right)+\dots+\lambda^{i_k-i_1}\rho\left(\frac{a_{i_k}z^{i_k}}{a_{i_1}z^{i_1}}\right)\neq 0$ where $\rho:\mathcal{O}_K\to \mathcal{O}_K/\mathfrak{m}_K$ is the
reduction map. Then $z'=[\lambda]z$ will do the job.
We can now prove the statement: let $z\in K$ be any element in the annulus $r_1\leq |z|\leq r_2$. There exists $k\in\bZ$ such that $|f(z)|\leq |a_kz^k|$. Assume that $k\geq 0$. Then $|a_kz^k|\leq |a_k|r_2^k$. Using the lemma, find $u$ with $|u|=r_2$ such that $|f(u)|=|a_nu^n|$ and $|a_nu^n|\geq |a_mu^m|=|a_m|r_2^m$ for every $m$. It follows that $|f(u)|\geq |a_k|r_2^k\geq |f(z)|$. If $k<0$, then arguing in the same way with $r_2$ replaced by $r_1$ gives the result.
