Not quite; you must still consider why the original contradiction $R \in R \iff R \not \in R$ no longer applies.
That is because once you limit to $B$, you can no longer conclude $R \in R \iff R \not \in R$ anymore; you can only conclude that under the additional assumption that $R \in B$. So the conclusion is simply that $R \not \in B$, and there is no contradiction.
Note that this holds even if you remove the axiom of regularity.
Here is what the author means by a "universal set" (see p. 314):
Most mathematical discussions are carried on within some context. For example, in a certain situation all sets being considered might be sets of real numbers. In such a situation, the set of real numbers would be called a universal set or a universe of discourse for the discussion.
Everything what is done in sequel only concerns subsets of such a universal set $U$ (union, intersection, ...).
This does no mean that one universal set suffices for all purposes. For example, if $U = \mathbb R$, then it is clear that there are also releavant sets which are no subsets of $U$ (for example $\mathbb R^n$ with $n > 1$).
When the author introduces the power set $\mathscr P(A)$ of a set $A$ and the Cartesian product $A_1 \times \ldots \times A_n$ of sets $A_1,\ldots, A_n$, it becomes very clear that certain natural constructions with sets produce new sets not belonging to a given universe $U$. For example, $\mathscr P(U)$ is never a subset of $U$.
What the author writes about Russell’s paradox is, in my opinion, misleading.
Let $U$ be a universal set and suppose that all sets under discussion are subsets of $U$.
It is not at all clear what this "suppose" means. Anyway, one can define the set
$$S = \{A | A ⊆ U \text{ and } A \notin A\}$$
This is a well-defined subset of $\mathscr P(U)$. Since $\mathscr P(U)$ is no subset of $U$, we cannot expect that a given subset of $\mathscr P(U)$ is a subset of $U$ - but it is not impossible. If it were true that the above $S$ is a subset of $U$, we would again end with the contradiction that $S \in S$ implies $S \notin S$ and $S \notin S$ implies $S \in S$ (Russell’s paradox). Therefore we conclude that $S$ is no subset of $U$, and there is no contradiction.
Thus "suppose that all sets under discussion are subsets of $U$" is certainly intended to say the following:
As long as we stay in $U$, it is impossible to reproduce Russell’s paradox. This paradox is a consequence of a "naive", but illegitimate definition of the "set" of all sets $A$ with $A \notin A$.
Best Answer
Cantor's theorem is a theorem, not a paradox.
Russel's paradox is also not a real paradox, but really a very short and elegant proof that the class of all sets is not a set.
The proof of Cantor's theorem uses a very similar idea as that of Russel's. This is not so surprising, as the conclusions are also related. It is very easy to deduce from Cantor's theorem that the class of all sets is not a set.
I hope this helps clearing the confusion.