Locally compact Stone duality: Can a boolean space be recovered from its boolean algebra of clopen sets alone?

Not in general: two non-isomorphic boolean spaces can have isomorphic algebras of clopen sets. Indeed, let $X$ be any boolean space that is not compact, and let $B=Cl(X)$. Then the Stone space $Y$ of $B$ is a boolean space with $Cl(Y)\cong Cl(X)$, but $Y$ is not homeomorphic to $X$ since $Y$ is compact and $X$ is not.

However, if you restrict to second-countable spaces, then $X$ can be reconstructed from $Cl(X)$. Indeed, the compact open sets can be identified as those elements $b\in Cl(X)$ such that $\{c\in Cl(X):c\leq b\}$ is countable. From there, the locally compact version of Stone duality can be used to reconstruct $X$.

To prove this characterization of the compact open sets, note that $\{c\in Cl(X):c\leq b\}$ is just the clopen algebra $Cl(b)$, considering $b$ as a subspace of $X$. So replacing $X$ with $b$, it suffices to show that if $X$ is a second-countable boolean space, then $X$ is compact iff $Cl(X)$ is countable. To prove this, choose a countable basis $(B_n)_{n\in\mathbb{N}}$ for $X$ consisting of compact open sets. If $X$ is compact, every clopen set is a finite union of basis elements (by compactness), and so $Cl(X)$ is countable. If $X$ is not compact, then for each $k\in\mathbb{N}$, $C_k=\bigcup_{n<k}B_n$ is compact and thus a proper subset of $X$. By induction, we can choose a sequence of disjoint nonempty compact open subsets $D_k\subset X\setminus C_k$ for each $k$. Then for every $A\subseteq\mathbb{N}$, $\bigcup_{k\in A} D_k$ is clopen, giving uncountably many distinct elements of $Cl(X)$. (The assumption that $D_k\subset X\setminus C_k$ guarantees that $\bigcup_{k\in A}D_k$ is closed, since its intersection with each $C_k$ is just a finite union and thus closed, and the $C_k$ are an open cover of $X$.)

(Alternatively, using the additional assumption that $X$ has no isolated points, you can just cite the classification in Henno Brandsma's answer, and so you just have to check that $Cl(C)$ is countable and $Cl(C\setminus\{0\})$ is uncountable.)

If you know that $X$ is locally compact (but not compact) zero-dimensional Hausdorff and moreover second-countable and crowded (no isolated points) then $X$ is homeomorphic to $C \setminus \{0\}$ (the Cantor set minus a point; which point does not matter). The proof follows easily from the observation that the one-point compactification of such an $X$ is homeomorphic to $C$, by its classical characterisation of $C$ as up to homeomorphism the only crowded compact metrisable zero-dimensional space.

So topologically speaking you're only interested in two spaces: $C$ (compact case) or $C$ minus a point (the non-compact case).