$\newcommand{\al}{\alpha}$In [leo monsaingeon's answer], for the the value $J(\al)$ of the infimum it was shown that 
\begin{equation*}
	J(\al)\le9
\end{equation*}
and conjectured that 
\begin{equation*}
	J(\al)\to9 
\end{equation*}
as $\al\downarrow0$. 

This is to confirm the conjecture and, moreover, to provide an explicit lower bound on $J(\al)$:  
\begin{equation*}
	J(\al)\ge9(1-\al)-\tfrac{3\pi}2\,(32\pi\al)^{1/3}. \tag{$*$}\label{*}  
\end{equation*}

Let 
\begin{equation*}
	\text{$a:=\pi/3$ and $b:=2\pi/3$.}
\end{equation*}

>**Lemma 1:** If $f$ is absolutely continuous on $[a,b]$ with $\int_a^b f^2\le\al$, then 
\begin{equation*}
	\int_a^b f'^2\ge\frac{(f_a+f_b)^4}{64\al}. 
\end{equation*}
Here and in what follows,
\begin{equation*}
	\text{$f_a:=|f(a)|$ and $f_b:=|f(b)|$.}
\end{equation*}

>**Lemma 2:** If $f$ is absolutely continuous on $[0,a]$ with $\int_0^a f^2\ge c^2$ for some real $c\ge0$, then 
\begin{equation*}
	\int_0^a f'^2\ge9(c-f_a\sqrt{a/3})_+^2\ge9(c^2-2cf_a\sqrt{a/3}), 
\end{equation*}
where $u_+:=\max(0,u)$. 


>**Lemma 2':** If $f$ is absolutely continuous on $[b,\pi]$ with $\int_b^\pi f^2\ge 1-\al-c^2$ for some $c\in[0,\sqrt{1-\al}]$, then 
\begin{equation*}
	\int_b^\pi f'^2\ge9(1-\al-c^2-2\sqrt{1-\al-c^2}f_b\sqrt{(\pi-b)/3}).  
\end{equation*}

It follows from these lemmas that 
\begin{equation*}
	\int_0^\pi f'^2\ge B(u):=9(1-\al)+\frac{u^4}{64\al}-2\pi u,  
\end{equation*}
where $u:=f_a+f_b$. Minimizing $B(u)$ in $u$, we get \eqref{*}. 

It remains to prove the lemmas, which will be done later. 

  [1]: https://mathoverflow.net/a/414597/36721