$\newcommand{\si}{\sigma}\newcommand{\R}{\mathbb R}\newcommand{\pa}{\partial}$The answer is no. The idea is to get a diffusion version of my [two-state Markov chain example][1]. 

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][2] 
\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 $t\ge1$. $\quad\Box$ 

[1]: https://mathoverflow.net/questions/447513/does-the-entropy-of-a-sde-with-nondegenerate-noise-always-increase#comment1156881_447513 
[2]: https://en.wikipedia.org/wiki/Fokker%E2%80%93Planck_equation