$\newcommand{\al}{\alpha}$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*}

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.

$\newcommand{\al}{\alpha}$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 post is to confirm the conjecture and, moreover, to provide an explicit lower bound on $J(\al)$:
\begin{equation*} J(\al)\ge9(1-\al-3^{1/3} (2 \pi)^{2/3}\al^{1/3}) \tag{$*$}\label{*} \end{equation*} for $\al\in(0,1]$.

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} \tag{$\clubsuit$}\label{-2} \end{equation*} or \begin{equation*} f_a+f_b\le\sqrt{\frac{16\al}{b-a}}. \tag{$\clubsuit\clubsuit$}\label{-1} \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 ($f(0)=0$ and) $\int_0^a f^2\ge c^2$ for some $c\in[0,1]$, then \begin{equation*} \int_0^a f'^2\ge9(c-f_a\sqrt{a/3})_+^2\ge9(c^2-2f_a\sqrt{a/3}), \tag{$\heartsuit$}\label{-0.5} \end{equation*} where $u_+:=\max(0,u)$.

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

These lemmas will be proved at the end of this answer.

In the case when \eqref{-2} holds, it follows from Lemmas 2 and 2' and the definitions $a:=\pi/3$ and $b:=2\pi/3$ that \begin{equation*} \int_0^\pi f'^2=\int_a^b f'^2+\int_0^a f'^2+\int_b^\pi f'^2 \ge B_1(u):=9(1-\al)+\frac{u^4}{64\al}-6\sqrt\pi\, u, \end{equation*} where $u:=f_a+f_b$. Minimizing $B_1(u)$ in $u$, we get \eqref{*}.

In the remaining case, by Lemma 1, \eqref{-1} must hold. So, it then follows from Lemmas 2 and 2' that \begin{equation*} \int_0^\pi f'^2\ge\int_0^a f'^2+\int_b^\pi f'^2 \ge 9(1-\al-2(f_a+f_b)\sqrt\pi/3) \ge 9(1-\al-8\sqrt{\al/3}). \end{equation*} For $\al\in(0,1]$, the latter expression is no less than the expression on the right-hand side of \eqref{*}.

Thus, \eqref{*} follows from Lemmas 1, 2, and 2'.

It remains to prove these lemmas.

Proof of Lemma 1: One can see that a minimizer $f$ of $\int_a^b f'^2$ subject to the conditions on $\int_a^b f^2$, $|f(a)|$, and $|f(b)|$ exists and is a real-analytic function on $[a,b]$. Replacing such a minimizer $f$ by $|f|$ does not change the values of $\int_a^b f'^2$, $\int_a^b f^2$, $|f(a)|$, and $|f(b)|$. So, without loss of generality (wlog) we may assume that $f$ is nonnegative minimizer. Let $c\in[a,b]$ be a point of minimum of $f$ on $[a,b]$, so that $0\le f(c)\le f(x)$ for all $x\in[a,b]$.

Note that no point $\xi$ in the interval $(a,c)$ can be the point of strict local maximum of $f$: Otherwise, if $m:=\max(f(\xi_1),f(\xi_2))<f(\xi)$ for some $\xi_1$ and $\xi_2$ such that $a<\xi_1<\xi<\xi_2<c$, take any $l\in(m,f(\xi))$ and get a new function $g$ by replacing $f(x)$ by $g(x):=\min(l,f(x))$ for all $x\in[\xi_1,\xi_2]$, with $g=f$ on $[a,b]\setminus[\xi_1,\xi_2]$. Then the conditions on $\int_a^b f^2$, $|f(a)|$, and $|f(b)|$ will hold with $g$ in place of $f$, whereas $\int_a^b g'^2$ will be strictly less than $\int_a^b f'^2$ -- which will contradict the assumption that $f$ is a minimizer.

Therefore and because $c$ is a point of (global) minimum of $f$ on $[a,b]$, it now follows that no point $\eta$ in the interval $(a,c)$ can be the point of strict local minimum of $f$: otherwise, there would exist a point in $(\eta,c)\subseteq(a,c)$ of strict local maximum of $f$.

So, $f$ is decreasing on $[a,c]$ and, similarly, increasing on $[c,b]$. Hence, for any real \begin{equation*} h\ge f_c:=f(c), \tag{0}\label{0} \end{equation*} the set \begin{equation*} \{x\in[a,b]\colon f(x)>h\} \end{equation*} is the complement to $[a,b]$ of an interval $[a_h,b_h]\subseteq[a,b]$ such that
\begin{equation*} \text{(i) $f(a_h)=h$ or $a_h=a\quad$ and $\quad$ (ii) $f(b_h)=h$ or $b_h=b$.} \end{equation*} Next, \begin{equation*} \al\ge\int_a^b f^2\ge\int_a^{a_h} f^2+\int_{b_h}^b f^2 \ge(a_h-a+b-b_h)h^2, \end{equation*} so that \begin{equation*} (a_h-a)+(b-b_h)\le\frac\al{h^2}. \tag{1}\label{1} \end{equation*} By Cauchy--Schwarz inequality, \begin{equation*} (a_h-a)\int_a^{a_h} f'^2\ge\Big(\int_a^{a_h} f'\Big)^2=(f_a-h)_+^2 \end{equation*} and similarly $(b-b_h)\int_{b_h}^b f'^2\ge(f_b-h)_+^2$. So, in view of \eqref{1}, \begin{equation*} \int_a^b f'^2\ge\int_a^{a_h} f'^2+\int_{b_h}^b f'^2 \ge\frac{(f_a-h)_+^2}u+\frac{(f_b-h)_+^2}{\al/h^2-u}, \tag{2}\label{2} \end{equation*} where $u:=a_h-a$; if the denominator of either one of the ratios in \eqref{2} is $0$, then the corresponding ratio is understood as $0$. Minimizing the sum of the ratios in \eqref{2} in $u\in[0,\al/h^2]$, we get \begin{equation*} \int_a^b f'^2\ge\frac{h^2}\al\,((f_a-h)_+ + (f_b-h)_+)^2 \ge\frac{h^2}\al\,((f_a+f_b-2h)_+)^2. \tag{3}\label{3} \end{equation*}

Now we need to distinguish between two cases:

Case 1: $f_a+f_b\ge4f_c$ and

Case 2: $f_a+f_b<4f_c$.

In Case 1, condition \eqref{0} will be satisfied with $h_*:=(f_a+f_b)/4$ in place of $h$. So, substituting $h_*$ for $h$ in \eqref{3}, we get \eqref{-2}.

In Case 2, write $f_c^2(b-a)\le\int_a^b f^2\le\al$. So, \eqref{-1} follows by the case condition.

This completes the proof of Lemma 1. $\quad\Box$

Proof of Lemma 2: Wlog, $f(a)\ge0$, so that $f(a)=f_a$. Let \begin{equation*} g(t):=f(t)-kt, \end{equation*} where \begin{equation*} k:=\frac{f_a}a, \end{equation*} so that $g(0)=0=g(a)$ and hence, by Wirtinger's inequality, \begin{equation*} \int_0^a g'^2\ge\frac{\pi^2}{a^2}\,\int_0^a g^2. \end{equation*} Also, in view of the fact that $\int_0^a g'=g(a)-g(0)=0$, we have \begin{equation*} \int_0^a f'^2\ge\int_0^a (g'+k)^2=\int_0^a g'^2+k^2 a\ge\int_0^a g'^2. \end{equation*} So, \begin{equation*} \int_0^a f'^2\ge\frac{\pi^2}{a^2}\,\int_0^a g^2=9\int_0^a g^2. \tag{4}\label{4} \end{equation*} By Minkowski's inequality, \begin{equation*} c\le\sqrt{\int_0^a f^2}\le\sqrt{\int_0^a g^2}+k\sqrt{\int_0^a t^2\,dt}, \end{equation*} whence \begin{equation*} \int_0^a g^2\ge(c-f_a\sqrt{a/3})_+^2. \end{equation*} Now the first inequality in \eqref{-0.5} follows from \eqref{4}, and the second inequality in \eqref{-0.5} is trivial.

This completes the proof of Lemma 2. $\quad\Box$

Proof of Lemma 2': Lemma 2' is quite similar to Lemma 2: it is obtained from Lemma 2 by the reflexions $t\leftrightarrow\pi-t$ and $c\leftrightarrow\sqrt{1-\al-c^2}$. $\quad\Box$