The Borel $\sigma$-algebras may be unequal.

Let $\omega_1$ be the least uncountable ordinal. Let $L_0=[0,\omega_1)\times I/\sim$ be the "long line", where $[0,\omega_1)$ is the space of ordinals less than $\omega_1$ with the order topology, $I=[0,1]$ is the unit interval, and the identifications are: $(\alpha,1)\sim(\alpha+1,0)$. Let $L'=L_0\cup\{*\}$ be the one point compactification of $L_0$.

Such $L'$ is a connected, linearly ordered compact space. Note that the Stone-Cech compactification of $L_0$ is again $L'$. The subspace $\omega_1=[0,\omega_1)\times\{0\}\subseteq L'$ is not Borel. Also for every closed subset $A\subseteq L_0$ that is not bounded above, every continuous function $f:A\to\mathbb{R}$ is eventually constant.

Let $L$ be a refinement of $L'$: a subset $U$ is open in $L$ if it is open in $L'$ or it is of the form $V\setminus\omega_1$ where $V$ is open in $L'$. Here $\omega_1$ is a closed hence a Borel subset of $L$. But if $f:L\to\mathbb{R}$ is continuous then it has to be eventually constant then it is continuous as a function on $L'$. Thus $C_b(L')=C_b(L)$, but $\mathcal{B}(L')\subsetneq\mathcal{B}(L)$.

The Borel $\sigma$-algebras may be unequal.

Let $\omega_1$ be the least uncountable ordinal. Let $L_0=[0,\omega_1)\times I/\sim$ be the "long line", where $[0,\omega_1)$ is the space of ordinals less than $\omega_1$ with the order topology, $I=[0,1]$ is the unit interval, and the identifications are: $(\alpha,1)\sim(\alpha+1,0)$. Let $L'=L_0\cup\{*\}$ be the one point compactification of $L_0$.

Such $L'$ is a connected, linearly ordered compact space. Note that the Stone-Cech compactification of $L_0$ is again $L'$. The subspace $\omega_1=[0,\omega_1)\times\{0\}\subseteq L'$ is not Borel. Also for every closed subset $A\subseteq L_0$ that is not bounded above, every continuous function $f:A\to\mathbb{R}$ is eventually constant.

Let $L$ be a refinement of $L'$: a subset $U$ is open in $L$ if it is of the form $U=V\cup (W\setminus\omega_1)$ where $V,W$ are open in $L'$. Here $\omega_1$ is a closed hence a Borel subset of $L$. But if $f:L\to\mathbb{R}$ is continuous then it has to be eventually constant then it is continuous as a function on $L'$. Thus $C_b(L')=C_b(L)$, but $\mathcal{B}(L')\subsetneq\mathcal{B}(L)$.