$\newcommand{\R}{\mathbb R}\newcommand{\Z}{\mathbb Z}\newcommand{\de}{\delta}\newcommand{\al}{\alpha}$The answer is no. In fact, $I_t$ does not have to be small even for small $t>0$.
Indeed, take any $\al\in(0,1)$ and suppose that $d=1$,
\begin{equation*}
f=\sum_{j\in\Z}(-1)^j\,1_{[j\de,(j+1)\de)} \tag{10}\label{10}
\end{equation*}
for some real $\de>0$, and $\ell$ is the standard normal density, Then all your conditions hold for some universal constant $c>0$.
Take any $t\in(0,1)$ and note that
\begin{equation*}
I_t=\int_\R dz\,(1+|z|)E|f(z+hZ)-f(z)|\ell(z+hZ),
\end{equation*}
where
\begin{equation*}
h:=\sqrt{2t}
\end{equation*}
and $Z\sim N(0,1)$.
Let
\begin{equation*}
J_t:=\int_\R dz\,(1+|z|)H(z)\ell(z),
\end{equation*}
where
\begin{equation}
H(z):=E|f(z+hZ)-f(z)|.
\end{equation}
Since $|f|=1$, we get
\begin{equation*}
|I_t-J_t|\le K_t:=2\int_\R dz\,(1+|z|)E|\ell(z+hZ)-\ell(z)|=2(K_{1,t}+K_{2,t}),
\end{equation*}
where
\begin{equation*}
K_{1,t}:=\int_\R dz\,(1+|z|)E|\ell(z+hZ)-\ell(z)|\,1(h|Z|\le1+|z|/2),
\end{equation*}
\begin{equation*}
K_{2,t}:=\int_\R dz\,(1+|z|)E|\ell(z+hZ)-\ell(z)|\,1(h|Z|>1+|z|/2).
\end{equation*}
Next,
\begin{equation*}
K_{1,t}\le\int_\R dz\,(1+|z|)Eh|Z|m(z)=Ch,
\end{equation*}
where $m(z):=\max\{|\ell'(u)|\colon|u|\ge|z|/2-1\}$. Here and in what follows, $C$ denotes various positive universal constants, possibly different even within one expression.
Next,
\begin{equation*}
K_{2,t}\le\int_\R dz\,(1+|z|)E1(h|Z|>1+|z|/2)\le C\sqrt h.
\end{equation*}
So,
\begin{equation*}
|I_t-J_t|\le C\sqrt h, \tag{20}\label{20}
\end{equation*}
Next, for any real $z$ with $f(z)=-1$,
\begin{equation*}
H(z)\ge2\sum_{m\in\Z}P(z+hZ\in[2m\de,(2m+1)\de))\to1 \tag{30}\label{30}
\end{equation*}
uniformly in $z$; here and in what follows, $\de\downarrow0$; \eqref{30} holds because $P(Z-\de\in B)-P(Z\in B)\to0$ uniformly over all Borel subsets $B$ of $\R$.
Using this, we similarly get
\begin{equation*}
J_t\ge\int_\R dz\,1(f(z)=-1)(1+|z|)(1-o(1))\ell(z)\to A:=\frac12\int_\R dz\,(1+|z|)\ell(z).
\end{equation*}
So, by \eqref{20},
\begin{equation*}
I_t\ge A-o(1)-C\sqrt h>A/2
\end{equation*}
eventually (that is, for all small enough $\de>0$) assuming that $t>0$ is small enough so that $C\sqrt h<A/3$. Assume also that $t>0$ is small enough so that $c_1 t^{\al/2}<A/2$. Then we get a contradiction with the desired inequality $I_t\le c_1 t^{\al/2}$. $\quad\Box$
