$\newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\varepsilon}$
Take any probability measures $P_0,P_1$ absolutely continuous with respect (w.r.) to $Q$. 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\,\|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", indeed $D(P||Q)$ is strongly convex in $P$ w.r. to the total variation norm.
We see that the lower bound on $\De$ does not depend on $Q$.
Proof of Theorem 1. Take indeed any $t\in(0,1)$. 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, 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
\begin{align*}
\de&:=(1-t)h(f_0)+th(f_1)-h(f_t) \\
&=(1-t)t\,(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)(1-s)\,ds \\
&=(1-t)t\,(f_1-f_0)^2\,
\int_0^1\Big(\frac t{f_{u_0(t,s)}}+\frac{1-t}{f_{u_1(t,s)}}\Big)(1-s)\,ds,
\end{align*}
where
$$u_j(t,s):=(1-s)t+js.$$
So,
\begin{equation}
\De=\int\de\,dQ=(1-t)t\,\int_0^1(1-s)\,ds\,[tI(u_0(t,s))+(1-t)I(u_1(t,s))], \tag{1}
\end{equation}
where
\begin{equation*}
I(u):=\int\frac{(f_1-f_0)^2}{f_u}\,dQ.
\end{equation*}
Next, take any $u\in(0,1)$. Then $P_1$ is absolutely continuous w.r. to $P_u$. Introducing
$g_u:=\dfrac{dP_1}{dP_u}=\dfrac{f_1}{f_u}$, we have
\begin{multline*}
I(u)=\frac1{(1-u)^2}\,\int\frac{(f_1-f_u)^2}{f_u}\,dQ
=\frac1{(1-u)^2}\,\int(g_u-1)^2\,dP_u \\
\ge\frac1{(1-u)^2}\,\Big(\int|g_u-1|\,dP_u\Big)^2
=\frac1{(1-u)^2}\,\|P_1-P_u\|^2=\|P_1-P_0\|^2. \tag{2}
\end{multline*}
Note also that for any $t\in(0,1)$ and $s\in(0,1)$ we have $u_0(t,s)\in(0,1)$ and $u_1(t,s)\in(0,1)$ and hence, by (2), $I(u_j(t,s))\ge\|P_1-P_0\|^2$ for $j=0,1$.
Now Theorem 1 follows by (1).
Remark. The constant factor $\frac12$ in the lower bound in Theorem 1 is the best possible one. Indeed, assuming that $P_1$ is absolutely continuous w.r. to $P_0$ and introducing $f:=\frac{dP_1}{dP_0}$, after some rather straightforward manipulations, we get \begin{equation} \De=\int k(t,f)\,dP_0, \tag{*} \end{equation} where $\De$ and $f=\frac{dP_1}{dP_0}$ areis 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.