A family of sets is really a set whose elements are sets. In ZFC, the Axiom of Union states (taken from Jech's **Set Theory**):

Axiom of Union. For any $X$ there exists a $Y = \bigcup X$.

That is:
$$\forall X\\ \exists Y\\ \forall u\\ \left( u\in Y \leftrightarrow \exists z(z\in X\wedge u\in z)\right).$$

So if you have a *family* of sets, this will play the role of $X$ in the axiom; the sets in the family are the sets $z$. Thus, you cannot have the cardinalities "grow so fast" that the union will not be a set; that can only occur if your collection of sets is not itself a set but a proper class to begin with (e.g., if you tried to take the "union" of the collection of all ordinals), whether the family is disjoint or not.

So now suppose you have a family $X$ of sets, and you want to consider "the" *disjoint* union of the elements of $X$ (up to a bijection, which are the isomorphisms in the category of sets). Using the Axiom Schema of Replacement, with the function $\mathbf{F}(x) = x\times\{x\}$, there exists a set $Y = \mathbf{F}[X] = \{\mathbf{F}(x)\mid x\in X\}$. Then $Y$ is a set, and the elements of $Y$ are pairwise disjoint: if $\mathbf{F}(x)\cap\mathbf{F}(y)\neq\emptyset$, then there exists $z\in x\times\{x\}$ such that $z\in y\times\{y\}$. So $z=(a,x)$ for some $a\in x$, and $z=(b,y)$ for some $b\in y$; hence $(a,x)=(b,y)$, so $x=y$. So let $Z=\cup Y$. Then $Z$ is a disjoint union of the sets in $X$, and is a set because it is a union of $Y$ (using the axiom of unions), which is itself a set by the Axiom of Replacement.

**Edit:** Actually, I'm being needlessly complicated in the last paragraph; there is no need to invoke the Axiom of Replacement. Given a family $X$ of sets, we can take $Y=(\cup X)\times X$ (which is a set by the Axioms of Union and Power sets, and the Axiom Schema of Separation) and then take $Z=\{ (a,b)\in Y\mid a\in b\}$. This set achieves the same objective, since for each $x\in X$, the subset $Z_x = \{(a,b)\in Z\mid b=x\}$ is bijectable with $x$, the set $Z$ is the union of the $Z_x$, and $Z_x\cap Z_y\neq \emptyset$ if and only if $x=y$.