No.

First of all, take note that for compact zero-dimensional spaces $X$, the $\sigma$-algebra generated by all clopen sets is precisely the $\sigma$-algebra of Baire sets (recall that in a completely regular space, the $\sigma$-algebra of Baire sets is the $\sigma$-algebra generated by collection of all zero sets). A space $X$ is said to be perfectly normal if it is normal and every closed set is a $G_{\delta}$-set.

**Theorem:** A compact Hausdorff space $X$ is perfectly normal if and only if every Borel set is a Baire set.

The above Theorem is stated in the measure theory textbook by Bogachev and other sources.

On the other hand, $\beta\Gamma$ is not perfectly normal since the singletons in $\beta\Gamma\setminus\Gamma$ are never $G_{\delta}$-sets in $\beta\Gamma$. To see this, suppose that $U$ is a non-principal ultrafilter on $\Gamma$. If $\{U\}$ were a $G_{\delta}$-set, then there would be a decreasing sequence of sets $(R_{n})_{n\in\omega}$ such that $\{U\}=\bigcap_{n\in\omega}\{V\in\beta\Gamma\mid R_{n}\in V\}$.

Let $S_{n}=R_{n}\setminus R_{n+1}$ for all $n\in\omega$. Let $S=\bigcup_{n}S_{2n}$. Without loss of generality, assume that $S^{c}\in U$. Let
$V$ be an ultrafilter that contains each set $R_{n}$ and which contains $S$. Then $V\neq U$, but $V\in\bigcap_{n\in\omega}\{V\in\beta\Gamma\mid R_{n}\in V\}$. This is a contradiction. Therefore, the singletons in $\beta\Gamma\setminus\Gamma$ are not $G_{\delta}$ sets in $\beta\Gamma$.

Another way of seeing that the Baire and the Borel sets do not coincide is by observing that there are only $2^{\aleph_{0}}$ Baire sets in $\beta\Gamma$ while there are $2^{2^{\aleph_{0}}}$ many Borel sets in $\beta\Gamma$.