Claim 1:
The divisibility rule for a number '$a$' to be divided by '$n$' is as follows. Express the number '$a$' in base '$n+1$'. Let '$s$' denote the sum of digits of '$a$' expressed in base '$n+1$'. Now $n|a \iff n|s$. More generally, $a \equiv s \pmod{n}$.
Example:
Before setting to prove this, we will see an example of this. Say we want to check if $13|611$. Express $611$ in base $14$.
$$611 = 3 \times 14^2 + 1 \times 14^1 + 9 \times 14^0 = (319)_{14}$$
where $(319)_{14}$ denotes that the decimal number $611$ expressed in base $14$. The sum of the digits $s = 3 + 1 + 9 = 13$. Clearly, $13|13$. Hence, $13|611$, which is indeed true since $611 = 13 \times 47$.
Proof:
The proof for this claim writes itself out. Let $a = (a_ma_{m-1} \ldots a_0)_{n+1}$, where $a_i$ are the digits of '$a$' in the base '$n+1$'.
$$a = a_m \times (n+1)^m + a_{m-1} \times (n+1)^{m-1} + \cdots + a_0$$
Now, note that
\begin{align}
n+1 & \equiv 1 \pmod n\\
(n+1)^k & \equiv 1 \pmod n \\
a_k \times (n+1)^k & \equiv a_k \pmod n
\end{align}
\begin{align}
a & = a_m \times (n+1)^m + a_{m-1} \times (n+1)^{m-1} + \cdots + a_0 \\
& \equiv (a_m + a_{m-1} \cdots + a_0) \pmod n\\
a & \equiv s \pmod n
\end{align}
Hence proved.
Claim 2:
The divisibility rule for a number '$a$' to be divided by '$n$' is as follows. Express the number '$a$' in base '$n-1$'. Let '$s$' denote the alternating sum of digits of '$a$' expressed in base '$n-1$' i.e. if $a = (a_ma_{m-1} \ldots a_0)_{n-1}$, $s = a_0 - a_1 + a_2 - \cdots + (-1)^{m-1}a_{m-1} + (-1)^m a_m$. Now $n|a$ if and only $n|s$. More generally, $a \equiv s \pmod{n}$.
Example:
Before setting to prove this, we will see an example of this. Say we want to check if $13|611$. Express $611$ in base $12$.
$$611 = 4 \times 12^2 + 2 \times 12^1 + B \times 12^0 = (42B)_{12}$$
where $(42B)_{14}$ denotes that the decimal number $611$ expressed in base $12$, $A$ stands for the tenth digit and $B$ stands for the eleventh digit.
The alternating sum of the digits $s = B_{12} - 2 + 4 = 13$. Clearly, $13|13$. Hence, $13|611$, which is indeed true since $611 = 13 \times 47$.
Proof:
The proof for this claim writes itself out just like the one above. Let $a = (a_ma_{m-1} \ldots a_0)_{n+1}$, where $a_i$ are the digits of '$a$' in the base '$n-1$'.
$$a = a_m \times (n-1)^m + a_{m-1} \times (n-1)^{m-1} + \cdots + a_0$$
Now, note that
\begin{align}
n-1 & \equiv (-1) \pmod n\\
(n-1)^k & \equiv (-1)^k \pmod n \\
a_k \times (n-1)^k & \equiv (-1)^k a_k \pmod n
\end{align}
\begin{align}
a & = a_m \times (n-1)^m + a_{m-1} \times (n-1)^{m-1} + \cdots + a_0 \\
& \equiv ((-1)^m a_m + (-1)^{m-1} a_{m-1} \cdots + a_0) \pmod n\\
a & \equiv s \pmod n
\end{align}
Hence proved.
Pros and Cons:
The one obvious advantage of the above divisibility rules is that it is a generalized divisibility rule that can be applied for any '$n$'.
However, the major disadvantage in these divisibility rules is that if a number is given in decimal system we need to first express the number in a different base. Expressing it in base $n-1$ or $n+1$ may turn out to be more expensive. (We might as well try direct division by $n$ instead of this procedure!).
However, if the number given is already expressed in base $n+1$ or $n-1$, then checking for divisibility becomes a trivial issue.
An interesting case is $n=101$.
The $3$-digit palindrome $aba$ is divisible by 101 iff $b=0$.
The $4$-digit palindrome $abba$ is divisible by 101 iff $a=b$.
The $5$-digit palindrome $abcba$ is divisible by 101 iff $c=2a$.
The $6$-digit palindrome $abccba$ is divisible by 101 iff $a+b=c$.
The $7$-digit palindrome $abcdcba$ is divisible by 101 iff $d=2b$.
The $8$-digit palindrome $abcddcba$ is divisible by 101 iff $a+d=b+c$.
The $9$-digit palindrome $abcdedcba$ is divisible by 101 iff $e = 2(c-a)$.
The $10$-digit palindrome $abcdeedcba$ is divisible by 101 iff $a+b+e=c+d$.
...
Best Answer
Yes, the test for divisibility by $11$ generalizes to arbitrary radix. Using modular arithmetic this amounts to simply evaluating a polynomial. Notice that radix notation has polynomial form. Namely in radix $\rm\:b\:$ the digit string $\rm\ d_n\ \cdots\ d_1\ d_0\ $ denotes $\rm\ d_n\ b^n +\:\cdots\: + d_1\ b + d_0\: =\ P(b),\ $ where $\rm\:\ P(x)\: =\ d_n\ x^n +\:\cdots\: d_1\ x + d_0\ $ is the polynomial associated to the string of digits.
Therefore $\rm\ mod\ b+1\ $ we infer $\rm\ b\equiv -1\ $ $\Rightarrow$ $\rm\ P(b)\equiv P(-1)\equiv d_0 - d_1 + d_2 - d_3 +\:\cdots\:.\: $ Similar modular reductions yield analogous divisibility tests, e.g. see here for casting out $91$'s.
If modular arithmetic is unfamiliar then one may proceed more simply as follows:
Factor Theorem $\rm\;\Rightarrow\; x-a\ |\ P(x)-P(a)\ $ so $\rm\ x = b,\ a= -1\ \Rightarrow\ b+1\ |\ P(b)-P(-1)\:.$
See also this post and various other posts on generalizations of casting out nines.