$\newcommand{\de}{\delta}
\newcommand{\De}{\Delta}
\newcommand{\ep}{\varepsilon}$

Take any probability measures $P_0,P_1$ absolutely continuous with respect (w.r.) to $Q$ and to each other. We shall prove the following:

>**Theorem 1.** For any $t\in[0,1]$,
\begin{align*}	\De&:=(1-t)H(P_0)+tH(P_1)-H(P_t) \\ 
%&\ge\frac{(1-t)t}2\,\Big[(1-t)\int\Big(\frac{dP_1}{dP_0}-1\Big)^2\,dP_0+t\int\Big(\frac{dP_0}{dP_1}-1\Big)^2\,dP_1\Big] \\
&
\ge\frac{(1-t)t}2\,\|P_1-P_0\|^2(1-\|P_1-P_0\|/2)
\ge\frac{(1-t)t}4\,\|P_1-P_0\|^2, 
\end{align*}
where $\|P_1-P_0\|:=\int|dP_1-dP_0|$ is the total variation norm of $P_1-P_0$, 
\begin{equation*}
	H(P):=D(P||Q)=\int \ln\frac{dP}{dQ}\,dP,
\end{equation*}
and, for any elements $C_0,C_1$ of a linear space, $C_t:=(1-t)C_0+tC_1$. 
Thus, by ["A third definition[8] for a strongly convex function"][2], indeed $D(P||Q)$ is strongly convex in $P$ w.r. to the total variation norm. 

We see that the lower bounds on $\De$ do not depend on $Q$. 

*Proof of Theorem 1.* Let $f_j:=\frac{dP_j}{dQ}$ for $j=0,1$, so that $f_t=\frac{dP_t}{dQ}$. By [Taylor's theorem with the integral form of the remainder][1], for $h(x):=x\ln x$ and $j=0,1$ we have
\begin{equation*}
	h(f_j)=h(f_t)+h'(f_t)(f_j-f_t)+(f_j-f_t)^2\int_0^1 h''((1-s)f_t+sf_j)(1-s)\,ds,
\end{equation*}
whence, in view of Jensen's inequality for the convex function $[0,\infty)\ni x\to1/x$ and the probability measure $\mu(ds):=2(1-s)\,ds$ on $[0,1]$,  
\begin{align*}
	\de&:=(1-t)h(f_0)+th(f_1)-h(f_t) \\ 
	&=\frac{(1-t)t}2\,(f_1-f_0)^2\,
	\int_0^1\Big(\frac t{(1-s)f_t+sf_0}+\frac{1-t}{(1-s)f_t+sf_1}\Big)2(1-s)\,ds \\ 
	&\ge\frac{(1-t)t}2\,(f_1-f_0)^2\,\Big(\frac tA+\frac{1-t}B\Big),
\end{align*}
where 
\begin{align*}
	A&:=\int_0^1[(1-s)f_t+sf_0]2(1-s)\,ds=f_{2t/3}, \\ 
	B&:=\int_0^1[(1-s)f_t+sf_1]2(1-s)\,ds=f_{(2t+1)/3}. 
\end{align*}
So,
\begin{equation*}
	\de\ge\frac{(1-t)t}2\,I(t,f)f_0, %(f_1-f_0)^2\,u(t),
	\tag{1}
\end{equation*}
where 
\begin{equation*}
	f:=\frac{f_1}{f_0}=\frac{dP_1}{dP_0}, 
\end{equation*}
\begin{multline*}
I(t,f):=\frac1{f_0}\,(f_1-f_0)^2\,\Big(\frac t{f_{2t/3}}+\frac{1-t}{f_{(2t+1)/3}}\Big)
=\frac{9 (f-1)^2 ((f-1) t+1)}{(2 (f-1) t+3) (2 f t+f-2 t+2)} \\ 
\ge J(f):=\frac{3(f-1)^2}{f+1+f\vee1}. 
%	u(t):=\frac1{f_{2t/3}}+\frac1{f_{(2t+1)/3}}
%	\ge(1-t)u(0)+tu(1) \\ 
%	=(1-t)\Big(\frac1{f_0}+\frac1{f_{1/3}}\Big)+t\Big(\frac1{f_{2/3}}+\frac1{f_1}\Big) 
%	>\frac{1-t}{f_0}+\frac t{f_1}, 
\tag{2}
\end{multline*}
Since $J''(f):=\frac{54}{(f+1+f\vee1)^3}>0$, the function $J$ is convex. So, for any real $c>0$  we have $J(f)\ge J(c)+J'(c)(f-c)$. Take now any $u\in(0,1)$ and then 
\begin{equation}
v:=\frac{4-u}{1+2u}.\tag{3}	
\end{equation}
Then $v>1$, 
$$a(u):=J(u)+J'(u)(1-u)=-\frac{9 (1-u)^2}{(2 + u)^2}=a(v),$$
and 
\begin{multline*}
	J(f)\ge[J(u)+J'(u)(1-u)]\vee[J(v)+J'(v)(1-v)] \\ 
	=a(u)+\frac{J'(u)+J'(v)}2\,(f-1)+\frac{J'(v)-J'(u)}2\,|f-1|. 
\end{multline*}
Also, $\int f\,dP_0=1=\int dP_0$ and 
$$\int|f-1|\,dP_0=\|P_1-P_0\|=:\ep.$$
So,
\begin{equation*}
	\int J(f)\,dP_0\ge a(u)+\frac{J'(v)-J'(u)}2\,\ep=:K(\ep,u), 
\end{equation*}
with $v$ as in (3). 
So, in view of (1) and (2), 
\begin{multline*}
	\De=\int\de\,dQ
	\ge\frac{(1-t)t}2\,\int I(t,f)f_0\,dQ
	=\frac{(1-t)t}2\,\int I(t,f)\,dP_0 \\ 
	\ge\frac{(1-t)t}2\,\int J(f)\,dP_0
	\ge\frac{(1-t)t}2\,K(\ep,u). 
\end{multline*}
Note that $K(\ep,u)$ is rational in $u$. The maximum, say $m(\ep)$, of $K(\ep,u)$ in $u\in(0,1)$ is a root of a polynomial (of degree $4$) whose coefficients are polynomials in $\ep$. Using then a computer algebra package, one verifies that $m(\ep)\ge\ep^2(1-\ep/2)$; details of calculations can be found in the [Mathematica notebook][3] or [its pdf image][4]. 
Thus, $\De\ge\frac{(1-t)t}2\,\ep^2(1-\ep/2)$; that is, the first inequality in Theorem 1 is proved. The second inequality there is trivial. 

Theorem 1 is now completely proved. 

*Remark.* The constant factor $\frac12$ in the first lower bound in Theorem 1 is the best possible one, at least for small values of $\|P_1-P_0\|$. Indeed, after some rather straightforward manipulations, we get 
\begin{equation}
	\De=\int k(t,f)\,dP_0, \tag{*}
\end{equation}
where $\De$ and $f=\frac{dP_1}{dP_0}$ are as before and 
\begin{equation}
	k(t,f):=t f \ln f-(1-t+t f)\ln(1-t+t f). 
\end{equation}
Take now any $h\in(0,1)$ and let $f$ take values $1-h,1+h$ each on a set of $P_0$-measure $1/2$, so that $\|P_1-P_0\|=h$. Then, in view of (*), for each $t\in(0,1)$, 
\begin{equation}
\De=\frac12\,k(t,1-h)+\frac12\,k(t,1+h)\sim \frac{(1-t)t}2\,h^2=\frac{(1-t)t}2\,\|P_1-P_0\|^2	
\end{equation}
as $h\downarrow0$, which confirms the optimality claim. 

[1]: https://www.math.umd.edu/~jmr/141/remainder.pdf 

[2]: https://en.wikipedia.org/wiki/Convex_function#Strongly_convex_functions

[3]: https://my.pcloud.com/publink/show?code=XZ63LW7ZAoR8BbeCCDyYyf0oLmh0SSdgxWvX

[4]: https://my.pcloud.com/publink/show?code=XZO3LW7Zw52J0KqKIj5KgwBBFWqS7kGKwOGX