Another reference for "Answer = Yes":
Chapter 12 of the Handbook of Boolean algebras has the title "The number of Boolean Algebras".
Don Monk, the author of this chapter, writes: "For almost all classes K of BAs which have been an object of intensive study, there are exactly $2^\kappa$ isomorphism types of members of K of each infinite power $\kappa$."
In particular, this is true for K=interval algebras. There are $2^\kappa$ many linear orders $L$ of cardinality $\kappa$ such that the corresponding interval algebras $Int(L)$ are pairwise non-isomorphic. (The elements of $Int(L)$ are the finite unions of intervals of $L$.)
Here is an explicit sketch of Monk's proof (which he calls "of folklore nature"):
- Call a BA $B$ "atomic" if you can find an atom below every positive element, and call $x\in B$ "atomless" if there is no atom below $x$.
- For any BA $B$, let $I(B)$ be the (possibly improper) ideal generated by the atoms together with the atomless elements. (An ideal is improper if it is equal to the whole BA)
- Write $D(B)$ (the "derivative" of $B$) for the Boolean algebra $B/I(B)$.
- Let $A_0=\omega$ be the linear order of natural numbers, and let $A_1=1+\eta+\omega$ be the linear order of the nonnegative rationals followed by a copy of the natural numbers. Then $Int(A_0)$ is atomic, and $Int(A_1)$ is not, so you have two nonisomorphic algebras.
- (In both these linear orders we call the smalles element $0$.)
- Check that both $I(Int(A_0))$ and $I(Int(A_1))$ are improper, i.e. the derivatives of $Int(A_0)$, $Int(A_1)$ are singletons. This fact is responsible for the "crucial point" below.
- Now for any $i,j\in \{0,1\}$ consider $A_{i}\times A_{j}$ with the "inverse" lexicographic order (first compare two second components in $A_{j}$, and only when they are equal compare the first components)
- $A_0\times A_0$ and $A_0\times A_1$ lead to atomic interval algebras, but $A_1\times A_0$ and $A_1\times A_1$ do not.
- A crucial point is that the first factor "disappears" when taking the derivative: $D(Int(A_i\times A_j))$ is naturally isomorphic to $Int(A_j)$, and therefore atomic iff $j=1$. So the four interval algebras derived from $A_i\times A_j$ are pairwise nonisomorphic.
- This proof idea can be generalized to infinite (weak) products. For notational simplicity I restrict the proof sketch to countably many factors.
- For any sequence $x=(x(0),x(1),\dots)\in 2^\omega$ let $A_x$ be the weak product $\prod_i^{\rm wk} A_{x(i)}$, i.e., the set of all functions $f$ defined on $\omega$ which satisfy $f(i)\in A_{x(i)}$ for all $i$, and $f(i)=0$ for almost all $i$.
- The set $A_x$ is countable, and linearly ordered as an "inverse" lexicographic product.
- The main work to do is to check that the $n$-th derivative of $Int(A_x)$ is naturally isomorphic to $Int(\prod_{i\ge n}^{\rm wk} A_{x(i)})$, and hence is atomic iff $x(i)=0$.
- This show that for different $x,y\in 2^\omega$ the Boolean algebras $Int(A_x)$ and $Int(A_y)$ are not isomorphic.
Essentially the same proof can be used for uncountable $\kappa$, where the result is probably more interesting. (I seem to recall that all countable BAs are interval algebras.)