Radius of convergence is the point at which the function ceases to be analytic: false for $k \neq \mathbb{C}$? Theorem: Let $U$ be an open set in $\mathbb{C}$, $f$ an analytic function on $U$, and $z_0 \in U$.  Then $f$ has a power series expansion centered at $z_0$.  If $r$ is a positive real number, and $U$ contains the disc of radius $r$ centered at $z_0$, then the radius of convergence of that power series is at least $r$.  
Any analytic continuation of $f$ to a larger open set $W$ containing $U$ is unique, so the same result holds with $U$ replaced by $W$.  In other words, the boundary of the disc of convergence of a local power series expansion of an analytic function is the point at which the given function ceases to be analytic.
The definition of an analytic function has a natural generalization to any topological field $k$ which is complete with respect to some absolute value (the main examples are $k = \mathbb{R}$ or a finite extension of $\mathbb{Q}_p$).  For $U$ an open set of $k^n$, an analytic function $f: U \rightarrow k$ is one which has a local power series expansion about every point of $U$.  This is defined in Serre, Lie Groups and Lie Algebras.
When $k = \mathbb{R}$, the theorem is false.  Let $U = k$, and define $f: k \rightarrow k$ by $f(x) = \frac{1}{1+x^2}$.  Interpreting $f$ as the restriction to $\mathbb{R}$ of a complex analytic function, we see that $f$ is analytic, and about $z_0 = 0$ has the power series expansion
$$1 - x^2 + x^4 - \cdots$$
which has radius of convergence $1$ (which follows from the Theorem and the fact that $\frac{1}{1+z^2}$ is meromorphic on $\mathbb{C}$ with singularities at $i, -i$.  
What about when $k$ is a finite extension of $\mathbb{Q}_p$?  Is the theorem still true?
 A: The problem with carrying over your theorem from $\mathbb{C}$ to say finite 
extensions of $\mathbb{Q}_p$ is that the notion of "ceases to be analytic"
cannot be carried over. In your definition of when a function ceases
to be analytic you are considering a maximal connected open set
on which a given analytic function is defined.
The natural analogue of $\mathbb{C}$ is $\mathbb{C}_p$, which is obtained by first taking
the algebraic closure of $\mathbb{Q}_p$ (which is not complete) and then
the completion of this algebraic closure, which can be proved to be
algebraically closed (see the book of Koblitz).
The $p$-adic valuation $v_p$ can be extended uniquely to $\mathbb{C}_p$
and we can endow $\mathbb{C}_p$ with the $p$-adic topology,
by taking as a basis the balls
$$B(a,r)=\{ x\in \mathbb{C}_p: v_p(x-a)>r\},$$for all $a\in \mathbb{C}_p$, $r\in \mathbb{Q}$.
With respect to this topology, every nonempty open set of $\mathbb{C}_p$
is disconnected.
The complement of $B(a,r)$, that is
$$B(a,r)^c=\{ x\in \mathbb{C}_p : v_p(x-a)\leq r\}$$
is also open. Indeed, if $b\in B(a,r)^c$,
then $B(b,r)$ is contained in $B(a,r)^c$ by the fact that
$$v_p(u+v)=\min (v_p(u),v_p(v))$$for all $u$, $v\in \mathbb{C}_p$ such that $v_p(u)$ is 
different from $v_p(v)$.
Now let $O$ be any nonempty open subset of $\mathbb{C}_p$.
Choose $a\in O$ and then $r$ such that $B(a,r)$ is strictly contained in $O$.
Then $O$ is the union of two disjoint nonempty open sets,
i.e. $B(a,r)$ and the intersection of $O$ with $B(a,r)^c$.
This shows that your definition of a function ceasing to be analytic
cannot be carried over to $\mathbb{C}_p$. The same reasoning applies if you take
a finite extension of $\mathbb{Q}_p$ instead of $\mathbb{C}_p$.
There are more strange phenomena.
For instance, consider the function $f$ on $\mathbb{C}_p$, given by
$f(x)=1$ if $v_p(x)>0$, and $f(x)=0$ if $v_p(x)<0$.
This is everywhere analytic on $\mathbb{C}_p$ in the sense that
it is locally given by a power series everywhere.
However, it cannot be defined by a single power series on all of $\mathbb{C}_p$.
In fact, it is because of the strange property of the $p$-adic topology
that every nonempty open subset of $\mathbb{C}_p$ is disconnected,
that the usual complex analysis cannot be carried over.

The problem is see in $\mathbb{C}_p$ is how to cover $|z| < 2$ by finitely many balls $|z - a| < 1$? This is how we prove in $\mathbb{C}$ that a function $f$ analytic on $|z| < 2$ and given by a power series on each of those finitely many balls is given by a power series on $|z| < 2$.

In $\mathbb{C}_p$, $|z|<2$ cannot be covered by finitely many balls $|z-a|<1$.
For suppose it could, say by balls$$|z-a_1|<1,\ldots, |z-a_r|<1.$$
Each ball $|z-a_i|<1$ with $|a_i|<1$ is already contained in the ball
$|z|<1$, by the ultrametric inequality.
So without loss of generality we may assume that $a_1=0$
and $|a_i|\geq 1$ for $i=2,\ldots,r$.
Now if $|z-a_i|<1$ for some $i=2,\ldots,r$ then $|z|=|a_i|$, again by the 
ultrametric inequality.
This shows that every $z$ with $|z|<2$ either satisfies $|z|<1$,
or $|z|$ is one of the values $|a_2|,\ldots,|a_r|$.
But this is impossible, for instance all numbers $z=p^{-c}$ with $c\in \mathbb{Q}$
and $1\leq p^c<2$ lie in $\mathbb{C}_p$, $|z|<2$ and have $|z|=p^c$.
So if we let $z$ run through all elements of $\mathbb{C}_p$ with $1\leq |z|<2$,
then $|z|$ assumes infinitely many different values.
But the main reason why the results from complex analysis
cannot be carried over from $\mathbb{C}$ to $\mathbb{C}_p$ is, as said, the
disconnectedness of $\mathbb{C}_p$. Here, instead of $\mathbb{C}_p$ one could take
any finite extension $k$ of $\mathbb{Q}_p$; there one encounters the same problem.
Recall that $\mathbb{C}$ is connected, meaning that a nonempty open subset of
$\mathbb{C}$ cannot be the union of two disjoint, nonempty open subsets.
But $\mathbb{C}_p$ or any finite extension $k$ of $\mathbb{Q}_p$ 
is the union of the disk
$|z|<1$ and the annulus $|z|\geq 1$ which are both open.
For instance, if $z$ is any element of the annulus, then so is
every $w$ with $|w-z|<1$, by the ultrametric inequality.
Hence if $z$ is any element of the annulus then there is an open
ball around it which is also contained in the annulus.
Now define the function $f$ by $f(z)=0$ if $|z|<1$, and $f(z)=1$
if $|z|\geq 1$.
This function is locally analytic everywhere.
But if $r>1$, then $f(z)$ cannot be given by a power series on
$|z|<r$. For first of all, such a power series should be nonconstant,
since $f$ assumes two different values on $|z|<r$; and secondly,
one can show that any nonconstant power series assumes infinitely
many different values, while $f$ assumes only two different values.
