$\newcommand{\ep}{\varepsilon}\newcommand{\R}{\mathbb R}\newcommand{\N}{\mathbb N}$The answer is yes. 

Indeed, let us follow the proof of Theorem 4.5.6 in Bogachev's book. Note that the measure $\mu$ is $\sigma$-finite on the set $X_0:=\bigcup_{n\in\N}f_n^{-1}(\R\setminus\{0\})$ and $\int_E f_n\,d\mu=\int_{X_0\cap E} f_n\,d\mu$ for all $n$ and all $E\in\Sigma$. So, without loss of generality $\mu$ is $\sigma$-finite. So, there is a $\Sigma$-measurable partition $(X_k)_{k\in\N}$ of $X$ such that $\mu(X_k)<\infty$ for all $k$. 

Let now $d\nu:=h\,d\mu$, where $h=a_k:=2^{-k}/(1+\mu(X_k))\in(0,\infty)$ on $X_k$, and let $g_n:=f_n/h$. Then $\nu$ is a finite measure and $\int_E f_n\,d\mu=\int_E g_n\,d\nu$ for all $E\in\Sigma$. By Theorem 4.5.6 in Bogachev's book, the $g_n$'s are uniformly integrable wrt to $\nu$; that is, 
\begin{equation}
	\lim_{M\to\infty}\sup_n\int_{|g_n|>M}|g_n|\,d\nu=0, 
\end{equation}
which can be rewritten as 
\begin{equation}
	\lim_{M\to\infty}\sup_n\int_{|f_n|>Mh}|f_n|\,d\mu=0.  
\end{equation}
Therefore and because $h>0$ and $\int h\,d\mu=\int d\nu<\infty$, we conclude that the $f_n$'s are uniformly integrable wrt to $\mu$. $\quad\Box$