$\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}$Instead of $\int_0^1|f-1|>2\ep$, let us write \begin{equation*} \ep:=\frac12\int_0^1|f-1|. \end{equation*} We want to show that \begin{equation*} \min\Big(\int_0^{1/4}f,\int_{3/4}^1 f\Big)\le\frac14-\frac\ep8. \end{equation*}
Without loss of generality (wlog), $\ep>0$. By approximation, wlog the function $f\in C^2[0,1]$ with $f''>0$ on $(0,1)$ and the set $\{x\in[0,1]\colon f(x)\ge1\}$ is an interval $[a,b]$ such that $0<a<b<1$, with $f(a)=f(b)=1$.
So, \begin{equation*} \ep=\ep_1+\ep_2, \end{equation*} where \begin{equation*} \ep_1:=\int_0^a(1-f),\quad\ep_2:=\int_b^1(1-f). \end{equation*} Let \begin{equation*} \de_1:=\frac14-\int_0^{1/4}f=\int_0^{1/4}(1-f),\quad \de_2:=\frac14-\int_{3/4}^1 f=\int_{3/4}^1(1-f). \end{equation*} It suffices to show that \begin{equation*} \max(\de_1,\de_2)\overset{\text{(?)}}\ge\ep/8. \tag{*} \end{equation*}
Our main tool here will be the following lemma.
Lemma 1: Suppose that $f\in C^2[0,1]$ with $g''>0$ on $(0,1)$ and $g(u)=0$ for some $u\in(0,1)$. Then the function $h\colon[0,1]\to\mathbb R$ defined by \begin{equation*} h(x):=\frac1{(x-u)^2}\,\int_u^x g \end{equation*} for $x\in[0,1]\setminus u$, with $h(u):=g'(u)/2$, is nonincreasing. Here, $\int_u^x:=-\int_x^u$ if $x<u$.
The proof of this lemma consists in a triple application of the special-case l'Hospital-type rule for monotonicity given by Proposition 4.1.
In particular, it follows from Lemma 1 (applied to $g:=f-1$) that \begin{equation*} b-a\ge1/2. \tag{0} \end{equation*} Indeed, suppose (0) is false. Then there are $l_1\in(0,a)$ and $l_2\in(0,1-b)$ such that $l_1+l_2=b-a$ (e.g., let $l_1:=\frac{b-a}{a+1-b}\,a$ and $l_1:=\frac{b-a}{a+1-b}\,(1-b)$). By Lemma 1 (it helps to draw a picture), \begin{equation*} \int_a^{a+l_1}(f-1)\le\frac{l_1^2}{a^2}\,\int_a^0(f-1)=\frac{l_1^2}{a^2}\,\int_0^a(1-f) =\frac{l_1^2}{a^2}\,\ep_1\le\ep_1, \tag{0.5} \end{equation*} so that $\int_a^{a+l_1}(f-1)\le\ep_1$, and the latter inequality is strict if $\ep_1>0$. Similarly, $\int_{a+l_1}^b(f-1)=\int_{b-l_2}^b(f-1)\le\ep_2$, and the latter inequality is strict if $\ep_2>0$. So, \begin{equation*} \ep=\int_a^b(f-1)=\int_a^{a+l_1}(f-1)+\int_{a+l_1}^b(f-1)\le\ep_1+\ep_2=\ep, \end{equation*} and the latter inequality is strict, since $\ep>0$ and hence either $\ep_1>0$ and $\ep_2>0$. Thus, we have $\ep<\ep$, a contradiction, which proves (0).
In view of (0) and by symmetry, it is enough to consider the following two cases: \begin{equation*} 0<a\le1/4,\quad 3/4\le b<1 \tag{i} \end{equation*} and \begin{equation*} 1/4<a\le1/2,\quad 3/4\le b<1 \tag{ii} \end{equation*}
Consider now case (i). Then, letting \begin{equation*} \eta_1:=\int_a^{1/4}(f-1),\quad\eta_2:=\int_{3/4}^b(f-1), \end{equation*} we have \begin{equation*} \de_1=\ep_1-\eta_1,\quad\de_2=\ep_2-\eta_2. \tag{1.5} \end{equation*} Next, for some $s\in[0,1/4]$ and $t\in[3/4,1]$, \begin{equation*} c:=\max_{[0,1/4]}f-1=f(s)-1\ge0,\quad d:=\max_{[3/4,1]}f-1=f(t)-1\ge0. \end{equation*} Then \begin{equation*} \eta_1\le c(1/4-a)\le c/4,\quad\eta_2\le d(b-3/4)\le d/4. \tag{2} \end{equation*} Also, by the convexity of $f$, assuming wlog that $c\le d$, for $x\in[1/4,3/4]$ we get \begin{equation*} f(x)-1\ge c+\frac{x-s}{t-s}\,(d-c)\ge c+\frac{x-1/4}{1-1/4}\,(d-c)=:\hat f(x), \tag{3} \end{equation*} so that \begin{equation*} \int_{1/4}^{3/4}(f-1)\ge\int_{1/4}^{3/4}\hat f=\frac c3+\frac d6\ge\frac{c+d}6, \end{equation*} whence \begin{equation*} \ep=\int_a^b(f-1)=\int_a^{1/4}(f-1)+\int_{1/4}^{3/4}(f-1)+\int_{3/4}^b(f-1) \ge\eta_1+(c+d)/6+\eta_2, \end{equation*} whence, by (1.5), $\de_1+\de_2\ge(c+d)/6$. Also, by (1.5) and (2), $\de_1+\de_2\ge\ep-(c+d)/4$. So, \begin{equation*} \max(\de_1,\de_2)\ge\frac12\,(\de_1+\de_2) \ge\frac12\,\max((c+d)/6,\ep-(c+d)/4)\ge\ep/5. \end{equation*} So, (*) holds if case (i) takes place.
Consider now case (ii). By Lemma 1 (cf. (0.5)), \begin{equation*} \ep=\int_a^b(f-1)\le\frac{(b-a)^2}{a^2}\,\int_0^a(1-f)\le\frac{(1-a)^2}{a^2}\,\int_0^a(1-f) =\frac{(1-a)^2}{a^2}\,\ep_1 \end{equation*} and, similarly, \begin{equation*} \ep_1-\de_1\le\ep_1\frac{(a-1/4)^2}{a^2}, \end{equation*} which implies \begin{equation*} \de_1\ge n(a)\ep_1\ge m(a)\ep, \tag{4} \end{equation*} where \begin{equation*} n(a):=1-\frac{(a-1/4)^2}{a^2},\quad m(a):=\frac{a^2}{(1-a)^2}n(a). \end{equation*} Next (cf. (3)), for $x\in[a,3/4]$, \begin{equation*} f(x)-1\ge f(a)-1+\frac{x-a}{t-a}\,(f(t)-f(a))=\frac{x-a}{t-a}\,d\ge \frac{x-a}{1-a}\,d, \end{equation*} whence \begin{equation*} \int_a^{3/4}(f-1)\ge\frac{(3/4-a)^2}{1-a}\,d/2\ge\frac{(3/4-a)^2}{1-a}\,2\eta_2, \end{equation*} by (2). So, \begin{equation*} \ep=\int_a^{3/4}(f-1)+\int_{3/4}^b(f-1)\ge k(a)\eta_2, \end{equation*} where \begin{equation*} k(a):=2\frac{(3/4-a)^2}{1-a}+1. \end{equation*} So, \begin{equation*} \de_2=\ep_2-\eta_2=\ep-\ep_1-\eta_2\ge\ep-\de_1/n(a)-\ep/k(a). \tag{5} \end{equation*}
Minimizing now $\max(\de_1,\de_2)$ over all $\de_1,\de_2,a$ such that $1/4\le a<1$, $\de_1\ge m(a)\ep$ (cf. (4)), and $\de_2\ge\ep-\de_1/n(a)-\ep/k(a)$, we get $\max(\de_1,\de_2)\ge\ep/q$ in case (ii), where $q=5.5489\dots$ is a certain algebraic number. This calculation takes about 0.27 sec with Mathematica:
So, (*) holds in case (ii) as well. $\Box$