Is there a notion of integral rigid spaces which mirrors the theory of integral schemes? For instance, let $B$ be an integral affinoid algebra over a non-Archimedean field $k$. Then are the affinoid algebras corresponding to affinoid open subsets of $Sp(B)$ integral? More generally, what is a suitable notion of a integral rigid analytic space?
Particularly interesting is to know whether, the rational subsets of $Sp(B)$ are integral. Since the rational open subsets are the analogues of the principle open subsets of an affine scheme, I would like to know if anybody knows any reference where the notion of integral rigid spaces have been discussed. Standard references don't seem to contain anything about it.
The reason that you are unlikely to find this notion appearing in the literature is that it is not nearly as well-behaved as it is in the algebraic situation.
Let us take our base definition of an integral finite type $k$-scheme (for some field $k$) as one which is reduced (i.e. the local rings are reduced) and irreducible. The reducedness condition is (roughly) OK in the rigid setting, but irreducibility is much, much more delicate.
Example: The closed disk $\{|x|\leqslant 1\}$ is not topologically irreducible. Indeed, this is equivalent to all non-empty open subsets being dense, but if $a$ is such that $|a|=1$, then $\{|x-a|<1\}$ and $\{|x|<1\}$ are disjoint open subsets. $\blacksquare$
Since any definition of 'irreducible' should contain the closed disk, we see that naive irreducibility is not a good notion. What is used instead is Zariski irreducibility -- a rigid space $X$ is Zariski irreducible if it cannot be written as a union of two disjoint Zariski closed subsets. This is then satisfied if $X$ is the closed unit disk.
The problem is that Zariski irreducibility is not preserved under passage to arbitrary possibly non-Zariski (e.g., affinoid) opens, and so doesn't behave as you would expect. The reason for this is that the topology of rigid spaces is much richer.
Example: Let $X=\mathrm{Spa}(\mathbb{Q}_p[x,y](y^2-x^3-x^2))$. Then, $X$ is irreducible, essentially because $\mathrm{Spec}(\mathbb{Q}_p[x,y]/(y^2-x^3-x^2))$ is. That said, consider the affinoid open $\{|x|\leqslant \tfrac{1}{2}\}$. Then, one can see that $$\sqrt{1+x}=1+\sum_{n\geqslant 1}{{\tfrac{1}{2}}\choose{n}}x^n,$$ converges $p$-adically on this affinoid, and thus $y^2-x^3-x^2$ is no longer Zariski irreducible. $\blacksquare$
Compare this with the notion of (geometrically) unibranch from scheme theory, where the above also gives a non-example. The non-unibranchness is detected by so-called 'analytic branches', which are something like irreducible components which appear 'analytically locally'. That is what is happening in the previous example, except in the analytic world 'analytically locally' is not a drastic operation, but the basic one!
Given this, a hint towards at least one good integral-like notion for rigid spaces is that of normal rigid spaces (compare with this). In particular, if $X$ is a normal rigid space, and $U\subseteq X$ is a connected affinoid open, then $\mathcal{O}(U)$ is an integral domain (cf.[1, Corollary 3.3.21]).
References:
1 Berkovich, V.G., 2012. Spectral theory and analytic geometry over non-Archimedean fields (No. 33). American Mathematical Soc..