I'd like to add that, after reading the preceeding answers, it becomes quite clear that $N : \mathcal{O}_K \rightarrow \mathbb{N}_{\geq 0}$, as defined in the second part of the question, is also bounded.

To see that, observe first that if

$$(a) = \prod_{i = 1}^r{\mathfrak{p}_i}^{e_i},$$

where none of the $\mathfrak{p}_i$ is principal and $r \geq (2h - 1)(h-1) + 1$, then $(a)$ will have at least two (essentially) distinct factorizations. Indeed, there exists an element in the class group of $K$, that is not the identity element, such that the corresponding equivalence class coincides with the equivalence classes of at least $2h$ of the factors of $(a)$, say $\mathfrak{p'}_1, ..., \mathfrak{p'}_{2h}$, whose exponents in the factorization of $(a)$ are $e'_1 \leq e'_2 ... \leq e'_{2h}$, respectively. Let $l$ denote the order of these ideals in the class group. Then the products $\mathfrak{p'}_1\mathfrak{p'}_2...\mathfrak{p'}_l, \mathfrak{p'}_1\mathfrak{p'}_{l+1}...\mathfrak{p'}_{2l}$ are principal, genrated by (essentially) distinct irreducible elements $a_1, a_2$. Note that $a_1^{e'_1}, a_2^{e'_1} | a$, but $a_1^{e'_1}a_2^{e'_1} \nmid a$. Thus $a$ has distinct factorizations (this is essentially Lior's argument, with a minor correction).

One can notice that, further, $N(a)$ must also be zero if $r \geq (2h - 1)(h-1) + 1 + 2h$, since, after selecting $\mathfrak{p'}_1, ..., \mathfrak{p'}_{2h}$ as above, one can form a principal ideal $(b) = \mathfrak{p'}_1 ...\mathfrak{p'}_{2h}$ that has distinct factorizations (for the same reason: $\mathfrak{p'}_1\mathfrak{p'}_2...\mathfrak{p'}_l = (b_1), \mathfrak{p'}_1\mathfrak{p'}_{l+1}...\mathfrak{p'}_{2l} = (b_2)$, the elements $b_1, b_2$ are distinct, irreducible, $b_1, b_2 | b$, but $b_1b_2 \nmid b$ ). Then $c = b^{-1}a$ will also have distinct factorizations, as the first observation implies. Hence $N(a) = 0$.

Thus we may restrict ourselves to the case $r \leq (2h - 1)(h - 1) + 2h$.
Let us call an irreducible algebraic integer absolutely irreducible if the corresponding principal ideal is a power of some prime ideal. If some $a$ now has "many" (that is, an arbitrarily large number) distinct factorizations, then there must exist an irreducible element $a'$ that is not absolutely irreducible and whose "large" power divides $a$ (since the number of distinct irreducible, nonassociate divisors of $a$ is now bounded in terms of $h$). One can then take $b$ to be a "large" power of $a'$, so that there also remains a "large" power of $a'$ to divide $c = b^{-1}a$. Then both $b, c$ will have distinct factorizations: one that contains as factors as many as possible $a'$'s, another - that contains as many as possible absolutely irreducible factors. Hence $N(a)$ is zero if the number of distinct factorizations of $a$ is large enough.

Non-Unique Factorizations: Algebraic, Combinatorial and Analytic Theory, Pure and Applied Mathematics278, Chapman & Hall/CRC, 2006. $\endgroup$