I think we established that the literature is lacking on this question. But I think the "correct" definition of morphisms between hyperstonean spaces can be puzzled together from G. Bezhanishvili's paper "Stone duality and Gleason covers through de Vries duality" (Topology and its Applications 157:1064-1080, 2010), especially section 6.
He proves in detail a duality between the category of complete Boolean algebras and complete Boolean algebra homomorphisms, and the category of extremally disconnected compact Hausdorff spaces and continuous open maps. But commutative von Neumann algebras and normal *-homomorphisms form a full subcategory of the former (via taking projections), which corresponds to the full subcategory of the latter consisting of hyperstonean spaces.
So Gelfand duality really restricts quite cleanly: commutative von Neumann algebras and normal *-homomorphisms are dual to hyperstonean spaces and open continuous maps.
No. Counterexamples include $X = \{0,1\}^\kappa$ or $[0,1]^\kappa$ for any $\aleph_0 < \kappa \le \mathfrak{c}$. These spaces are compact Hausdorff (Tikhonov's theorem) and are separable (Hewitt-Marczewski-Pondiczery).
(This special case of H-M-P can also be proved directly. For example, with $X =[0,1]^\kappa$, identify $\kappa$ with a subset of $[0,1]$, so that $X$ is a space of functions from $[0,1]$ to itself; then it is easy to show that the set of polynomials with rational coefficients is dense in $X$.)
Now the uncountable successor ordinal $\omega_1 + 1$, with its order topology, can be embedded in $X = \{0,1\}^\kappa$ or $X = [0,1]^\kappa$. See Fremlin, Measure Theory, 434K(d) (thanks to Robert Furber for the reference). One can also follow a construction similar to the Stone-Čech compactification and embed $\omega_1 + 1$ into $[0,1]^{C(\omega_1+1, [0,1])}$, noting that $|C(\omega_1 + 1, [0,1])| = \mathfrak{c}$. Either way, call the embedding $\phi$.
Let $\nu$ be the famous Dieudonné measure on $\omega_1 + 1$, where $\nu(B) = 1$ if $B$ contains a closed unbounded subset of $\omega_1$ and $\nu(B) = 0$ otherwise. (See Fremlin 411Q and 4A3J for details.) Then the pushforward $\mu = \nu \circ \phi^{-1}$ is a finite Borel measure on $X$ which is not regular.
Specifically, since $\phi$ is an embedding and $\omega_1$ is open in $\omega_1 + 1$, then the image $A = \phi(\omega_1)$ is a relatively open subset of the compact set $\phi(\omega_1 + 1)$, so that $A$ is the intersection of an open set and a closed set in $X$, and in particular is Borel. Moreover $\mu(A) = \nu(\omega_1) = 1$. But if $K$ is any compact subset of $A$, then $\phi^{-1}(K)$ is a compact subset of $\omega_1$, hence $\mu(K) = \nu(\phi^{-1}(K)) = 0$. Thus inner regularity fails.
This example is also Exercise 7.14.130 of Bogachev's Measure Theory.
Best Answer
If $L$ is a completely regular frame, then let $\mathfrak{R}(L)$ denote the point-free Stone-Cech compactification of $L$. Let $X$ be a compact Hausdorff space, and let $L$ be the frame of all open sets in $X$. Let $B$ be a complete Boolean algebra.
Let $Z$ denote the collection of all frame homomorphisms $\phi:L\rightarrow B$. Then $Z$ is a $B$-valued structure. If $L$ is regular and compact, then $V^B\models\text{$Z$ is a compact Hausdorff space}$, and $Z$ is the space that one would obtain by interpreting the space $X$ in the Boolean-valued forcing extension $V^B$ (when we interpret a compact Hausdorff space in a forcing extension, we usually need to add points to the space so that we retain compactness and we need to add open sets too).
Let $Z_1$ denote the collection of all frame homomorphisms $\phi:L\rightarrow\mathfrak{R}(B)$. Then by the universal property of the point-free Stone-Cech compactification, the set $Z_1$ can be put into a canonical one-to-one correspondence with $Z_1$. Let $Z_2$ be the collection of all continuous functions $f:S(B)\rightarrow X$. Then $Z_2$ can be put into a one-to-one correspondence with $Z_1$ and $Z$ as well. We have more correspondences that are closer to the idea of equivalence classes of measurable functions. The collection of frame homomorphisms from $L$ to $B$ corresponds to the collection of all localic maps from $B$ to $L$: If $f:L\rightarrow B$ is a frame homomorphism, then the corresponding localic map is the right Galois adjoint $f_*:B\rightarrow L$ defined by letting $f_*(b)=\bigvee\{x\mid f(x)\leq b\}$. The localic map $f_*$ preserves all infima.
Congruence tower
If $L$ is a frame, then the collection of all congruences on $L$ is another frame which we denote by $\mathfrak{C}(L)$. There is a homomorphism $\Delta:L\rightarrow\mathfrak{C}(L)$ where we set $(a,b)\in\Delta(x)$ precisely when $a\vee x=b\vee x$. We can iterate this process of taking congruence frames transfinitely and obtain $\mathfrak{C}^\alpha(L)$ for all ordinals $\alpha$ where we set $\mathfrak{C}^0(L)=L$ and $\mathfrak{C}^{\alpha+1}(L)=\mathfrak{C}(\mathfrak{C}^\alpha(L))$ and $\mathfrak{C}^\lambda(L)=\text{Colim}_{\alpha<\lambda}\mathfrak{C}^\alpha(L)$ for each limit ordinal $\lambda$. This transfinite process generally does not terminate. In each case, we have a canonical homomorphism $\Delta^\alpha:L\rightarrow\mathfrak{C}^\alpha(L).$
If $L$ is a frame and $B$ is a complete Boolean algebra, then for each homomorphism $\phi:L\rightarrow B$, there is a unique frame homomorphism $\overline{\phi}:\mathfrak{C}^\alpha(L)\rightarrow B$ where $\phi=\overline{\phi}\Delta^\alpha$. In particular, the frame homomorphisms from $L$ to $B$ can be put into a one-to-one correspondence with the frame homomorphisms from $\mathfrak{C}^\alpha(L)$ to $B$.
Baire functions induce our maps
Recall that the Loomis-Sikorski theorem states that every $\sigma$-complete Boolean algebra can be written as $\mathcal{M}/\mathcal{I}$ where $\mathcal{I}$ is a $\sigma$-ideal on a $\sigma$-algebra $(X,\mathcal{M})$, and I think it is quite natural to write a complete Boolean algebra as a
Suppose that $B$ is a complete Boolean algebra, and $B=\mathcal{M}/\mathcal{I}$ for some $\sigma$-algebra $(X,\mathcal{M})$ and $\sigma$-ideal $\mathcal{I}$. Let $K$ be a compact space, and let $f:X\rightarrow K$ be a Baire function. Then whenever $U$ is a co-zero set, we have $f^{-1}[U]\in\mathcal{M}$. Let $L$ be the lattice of open subsets of $B$, and let $G$ denote the sublattice of cozero subsets of $L$, and let $W$ denote the lattice of zero subsets of $L$. Define a mapping $f^\sharp:L\rightarrow\mathcal{M}/\mathcal{I}$ by letting $f^\sharp(U)=\bigvee\{f^{-1}[V]\oplus\mathcal{I}:V\in G,V\subseteq U\}$. It is not too hard to show that $f^\sharp(U)=\bigvee\{f^{-1}[V]\oplus\mathcal{I}:V\in G,\overline{V}\subseteq U\}=\bigvee\{f^{-1}[V]\oplus\mathcal{I}:C\in W,C\subseteq U\}.$
Suppose now that $C\in W,C\subseteq\bigcup_{\alpha\in A}U_\alpha$. Then by compactness, there are $\alpha_1,\dots,\alpha_r\in A$ where $C\subseteq U_{\alpha_1}\cup\dots\cup U_{\alpha_r}$. Each $U_{\alpha_i}$ is a union of countably many zero sets, so each set $C\cap U_{\alpha_i}$ is also the union of countably many zero sets $(C_{n,i})_{n=0}^\infty$. Therefore, $f^{-1}[C]\oplus\mathcal{I}=\bigvee_{i=1}^r\bigvee_{n=0}^\infty f^{-1}[C_{n,i}]\oplus\mathcal{I}\leq f^\sharp(U_{\alpha_1})\vee\dots\vee f^\sharp(U_{\alpha_n})\leq\bigvee_{\alpha\in A}f^\sharp(U_\alpha)$. Therefore, we conclude that $f^\sharp(\bigcup_{\alpha\in A}U_\alpha)=\bigvee\{f^{-1}[C]\oplus\mathcal{I}:C\in W,C\subseteq\bigcup_{\alpha\in A}U_\alpha\}\leq\bigvee_{\alpha\in A}f^\sharp(U_\alpha)$.
Therefore, $f^\sharp(\bigcup_{\alpha\in A}U_\alpha)=\bigvee_{\alpha\in A}f^\sharp(U_\alpha)$.
If $U,V\in L$, then $f^\sharp(U\cap V)=\bigvee\{f^{-1}[O]\oplus\mathcal{I}:O\in G,O\subseteq U\cap V\}$
$=\bigvee\{(f^{-1}[O_1]\cap f^{-1}[O_2])\oplus\mathcal{I}:O_1,O_2\in G,O_1\subseteq U,O_2\subseteq V\}$
$=\bigvee\{f^{-1}[O_1]\oplus\mathcal{I}:O_1\in G,O_1\subseteq U\}\wedge \bigvee\{f^{-1}[O_2]\oplus\mathcal{I}:O_2\in G,O_2\subseteq V\}$
$=f^\sharp(U)\wedge f^\sharp(V)$.
Therefore, $f^\sharp$ is a frame homomorphism.
On the other hand, the mapping $f\mapsto f^\sharp$ from Baire maps modulo $\mathcal{I}$ to frame homomorphisms is generally non-injective
I claim that we can have two Baire functions $f,g:X\rightarrow K$ where $f^\sharp=g^\sharp$ but where $f(x)\neq g(x)$ for all $x\in X$. Suppose that $K=\{0,1\}^\mathcal{I}$ and let $f:X\rightarrow K$ be the mapping where $f(x)=(\chi_A(x))_{A\in\mathcal{I}}$. We observe that the Baire sets in $K$ are generated by the sets $\pi_A^{-1}[\{1\}]$ where $\pi_A:\{0,1\}^\mathcal{I}\rightarrow\{0,1\}$ is the projection mapping. But $\pi_A\circ f=\chi_A$, so $f^{-1}[\pi_A^{-1}[\{1\}]]=\chi_A^{-1}[\{1\}]=A$, hence $f$ is Baire measurable. On the other hand, if $g:X\rightarrow\{0,1\}^\sharp$ is the zero function, then $f^\sharp(\pi_A^{-1}[\{1\}])=g^\sharp(\pi_A^{-1}[\{1\}])$ for all $A\in\mathcal{I}$, so $f^\sharp=g^\sharp$.