Not really a total answer, but as an addendum to Gerhard's comment, because there are only 10 digits which can appear there are several things which make the problem easier.
It's not immediately obvious, but there is a finite set of "minimal" partitions $(S_1,S_2)$ where each $S_i$ is a collection of digits. These minimal partitions have the following properties:
1). $sum(S_1) = sum(S_2)$
2). $sum(T_1) = sum(T_2)$ for $\emptyset\neq T_1\subseteq S_1$ and $\emptyset\neq T_2\subseteq S_2$ implies ($T_1 = S_1$ and $T_2 = S_2$)
Note that the sets $S_1 = S_2$ = $\lbrace d\rbrace$ are minimal by this definition.
Rough Outline of Proof: Condition 2 implies that a given digit $d$ cannot appear in both $S_1$ and $S_2$ and 0 does not appear in either one. Thus $S_1$ contains some digits and $S_2$ contains other digits (except for the single digit pairs noted above), and we can reduce to at most $3^9+10$ cases of which digits appear in $S_1$ and $S_2$ (for each digit, it either goes in $S_1$, in $S_2$, or in neither, plus the 10 single digit pairs).
Now lots of these cases get thrown away right away, for example if 1 and 2 are in $S_1$ and 3 is in $S_2$. Among the ones that remain, there are only finitely many ways to put in more copies of the digits to avoid equal subsums. \ \
Once one has this set of minimal partitions, one can use them as a sort of 'primes' for building up all larger partitions of a set of digits into equal subsums; every partition which has equal sums will have at least one of these minimal partitions as subsets of its digit sets, removing this subset one gets a strictly smaller set to deal with.
Using basic enumeration techniques from combinatorics, one can use them to give an upper bound on how many integers with a given number of digits can have such a partition. It is possible that there are multiple ways of decomposing a partition into these minimal sets, for example let n = 1123333444. Then there are at least two ways to decompose the digits of $n$ into minimal partitions:
$(\lbrace 1,1\rbrace,\lbrace 2\rbrace) \cup (\lbrace 3,3,3,3\rbrace,\lbrace 4,4,4\rbrace)$
$(\lbrace 1,1,4\rbrace,\lbrace 3,3\rbrace) \cup (\lbrace 2,3,3\rbrace,\lbrace 4,4\rbrace)$