$\newcommand{\si}{\sigma}\newcommand{\R}{\mathbb R}\newcommand{\pa}{\partial}$The answer is no. The idea is to get a diffusion version of my two-state Markov chain example.
Indeed, for $t\in(0,\infty)$ and real $x$, let \begin{equation} b(x,t):=e^{(t+1)^2 x^2/(2 t)}+\frac{1-t}{2 (t+1)^3}\ge1+\frac{1-t}{2 (t+1)^3}\ge\frac{53}{54}>\frac12, \end{equation} so that \begin{equation} \si(x,t):=\sqrt{2b(x,t)}\ge1. \end{equation} Moreover, letting \begin{equation} f_t(x):=f(x,t):=g_{0,t/(t+1)^2}(x), \end{equation} where $g_{a,s^2}$ is the density of the normal distribution with mean $a$ and variance $s^2$, we see that $f$ is a solution of the Fokker–Planck equation \begin{equation} \pa_t f(x,t)=\pa_x^2(b(x,t)f(x,t)). \end{equation} So, $f_t$ is the density of $X_t$ given the SDE \begin{equation} dX_t=\si(X_t,t)\,dW_t \end{equation} with the initial condition $X_0=0$ (since the $EX_t^2=t/(t+1)^2\to0$ as $t\downarrow0$).
However, the entropy \begin{equation} \int_\R f_t\ln\frac1{f_t}=\frac{1}{2} (\ln (2 \pi t)-2 \ln (t+1)+1) \end{equation} decreases in $t\ge1$. $\quad\Box$