For $0<\alpha<1/2$ derivative $u_x$ grows on the boundary (in absolute value) at least as $(1-t)^{\alpha-1/2}$.

To prove it I'll consider the domain $x>-(1-t)^\alpha$ instead of $x>(1-t)^\alpha$ in the question. It will simplify the exposition somewhat: not counting unnecessary minus signs, writing $x$ instead of $|x|$ etc.

A variable change $x\to x+(1-t)^\alpha$ turns the heat equation into
$$
Lu=u_t-\frac12u_{xx}-\alpha(1-t)^{\alpha-1}u_x=0
$$
and the domain into $D_+=\{x>0\}\times\{0<t<1\}$.

Let $Q_{\varepsilon,\delta}=(0,\varepsilon)\times(1-\delta,1)$, $0<\varepsilon,\delta\le1$ be a rectangle. It will be enough to prove the required estimate in such a domain.

First note that any solution $u$ for $f\not\equiv0$ on the parabolic boundary $PQ_{\varepsilon,\delta}$ of a small enough $Q_{\varepsilon,\delta}$ satisfy the following relations:
\begin{eqnarray*}
&&u(0,t)= 0, \\
C_1 x &\le& u(x,1-\delta)\le C_2x ,\\
C_3 &\le& u(\varepsilon,t)\le C_4,
\end{eqnarray*}
for some positive constants $C_i$.
If there is another solution $v$ with the same properties then there exist constants $K_1,K_2>0$ s.t. $K_1v\le u\le K_2v$ on $PQ_{\varepsilon,\delta}$ and consequently in $Q_{\varepsilon,\delta}$.

If function $v$ satisfy $Lv\le0$ instead of $Lv=0$ then
$K_1v\le u$ in $Q_{\varepsilon,\delta}$ and
$$
u_x(0,t)=\lim_{x\to0+}\frac{u(x,t)-u(x,0)}x =\lim_{x\to0+}\frac{u(x,t)}x
\ge K_1\lim_{x\to0+}\frac{v(x,t)}x=K_1v_x(0,t).
$$

Lets construct such a function $v$ with the required asymptotic of $v_x(0,t)$. Put $\psi(x)=|x|^{2\alpha}\,\mathrm{sign}\, x$, $Z(x,t)=(2\pi t)^{-1/2}\exp\{-x^2/2t\}$ and define
$$
v(x,t)=\int_{-\infty}^\infty Z(x-y,1-t)\psi(y)\,dy,
$$
a Poisson potential with reversed time.

This function loses its smoothness as $t\to1-0$ at the right rate, as can be checked by direct calculations:
$$
v_x(0,t)=\frac{2^{\alpha +\frac{1}{2}}\Gamma (\alpha +1) (1- t)^{\alpha -\frac{1}{2}}}{\sqrt{\pi }}.
$$

Since $v_t=-v_{xx}/2$ it is left to show that $Lv=-v_{xx}-\alpha(1-t)^{\alpha-1}v_x\le0$
in some $Q_{\varepsilon,\delta}$.

Denote $\tau=1-t$ and put
$$
w(x,\tau)=\int_{-\infty}^\infty Z(x-y,\tau)\psi(y)\,dy.
$$
Then $Lv=-w_{xx}-\alpha\tau^{\alpha-1}w_x$.
Derivatives present in the rhs don't change sign in $D_+$:
\begin{align*}
w_x(x,\tau)&=\int_{-\infty}^\infty Z(x-y,\tau)\psi'(y)\,dy>0, \\
w_{xx}(x,\tau)&=\int_{0}^\infty(Z(x-y,\tau)-Z(x+y,\tau))\psi''(y)\,dy<0.
\end{align*}

So one has to show that $\alpha\tau^{\alpha-1}w_x\ge|w_{xx}|$. To this end we'll get estimates from below for $w_x$ and from above for $|w_{xx}|$.

We will divide the domain $D_+$ into two parts, above and below the parabola $x=\tau^{1/2}$.

$ $1. $\{x\le\tau^{1/2}\}$.

To get estimates for $w_x$, $w_{xx}$ we'll use the fact that function $\psi$ is homogeneous. Namely, from $\psi(kx)=k^{2\alpha}\psi(x)$, $k>0$, it follows that $w(x,\tau)=k^{-2\alpha}w(kx,k^2\tau)$ and $w_x(x,\tau)=k^{1-2\alpha}w_x(kx,k^2\tau)$.
Putting $k=\tau^{1/2}$ gives
$$
w_x(x,\tau)=\tau^{\alpha-1/2}w_x(x\tau^{-1/2},1)
$$
and
$$
w_x(x,\tau)\ge \tau^{\alpha-1/2}\min_{0\le y\le1} w_x(y,1)=C\tau^{\alpha-1/2}
$$
for $0\le x\le \tau^{1/2}\le1$.
In the same manner, using the equality $w_{xx}(x,\tau)=k^{2-2\alpha}w_x(kx,k^2\tau)$ one gets
$$
|w_{xx}(x,\tau)|\le \tau^{\alpha-1}\max_{0\le y\le1} |w_{xx}(y,1)|=C\tau^{\alpha-1}.
$$
Now we can obtain the required inequality:
$$
Lv=-\alpha\tau^{\alpha-1}w_x-w_{xx}\le
-C_1 \tau^{\alpha-1} \tau^{\alpha-1/2}+ C_2\tau^{\alpha-1}=
C_1 \tau^{\alpha-1}(-\tau^{\alpha-1/2} + C_3)\le 0
$$
which holds for small enough values of $\tau$.

$ $2. $\{x>\tau^{1/2}\}$.

Preliminary we'll establish some asymptotic expansions for $\partial_x^k w(x,1)$, $x\to+\infty$.
To illustrate the idea, let's show that
$$
w(x,1)=x^{2\alpha}+O(x^{2\alpha -1}),\quad x\to+\infty.
$$

Cutting of the intervals with exponentially decreasing tails one gets
$$
w(x,1)=
\psi(x)+\int_{-\infty}^\infty Z(x-y,1)(\psi(y)-\psi(x))\,dy=
$$
$$
=\psi(x)+\int_{x/2}^{3x/2} Z(x-y,1)(\psi(y)-\psi(x))\,dy+O(e^{-x^2/16})=
$$
$$
=\psi(x)+\int_{x/2}^{3x/2} Z(x-y,1)(\psi(y)-\psi(x)-\psi'(x)(x-y))\,dy+O(e^{-x^2/16})=
$$
$$
=x^{2\alpha}+O(x^{2\alpha -1}),\quad x\to+\infty,
$$
since
$$
\left|\int_{x/2}^{3x/2} Z(x-y,1)(\psi(y)-\psi(x)-\psi'(x)(x-y))\,dy\right|\le
$$
$$
\le \frac12|\psi''(x/2)|\int_{-\infty }^{\infty } Z(x-y,1)(x-y)^2\,dy=Cx^{2\alpha -1}.
$$
Repeating this argument gives
\begin{equation}
\begin{split}
w_x(x,1)&=2\alpha x^{2\alpha-1}+O(x^{2\alpha -2}),\quad x\to+\infty, \\
w_{xx}(x,1)&=2\alpha(2\alpha-1) x^{2\alpha-2}+O(x^{2\alpha -3}),\quad x\to+\infty.
\end{split}
\end{equation}

Now return to the estimate of $Lv$.
From the positivity of $w_x(x,1)$ and
the established asymptotic expansion it follows that there exists $C>0$ s.t.
$$
w_x(x,1)\ge Cx^{2\alpha-1},\quad x\ge1.
$$
Inserting $k=\tau^{-1/2}$ in the equality $w_x(x,\tau)=k^{1-2\alpha}w_x(kx,k^2\tau)$, we have
$$
w_x(x,t)=\tau^{\alpha-1/2}w_x(x\tau^{-1/2},1)\ge C \tau^{\alpha-1/2} (x\tau^{-1/2})^{2\alpha-1}=Cx^{2\alpha-1}.
$$
In the same way one obtains
$$
|w_{xx}(x,\tau)|\le Cx^{2\alpha-2}.
$$
Finally
\begin{align*}
Lv&=-\alpha\tau^{\alpha-1}w_x-w_{xx}\le
-C_1x^{2\alpha-1}\tau^{\alpha-1}+C_2 x^{2\alpha-2}= \\
&=C_1x^{2\alpha-2}(-\tau^{\alpha-1/2}(x\tau^{-1/2})+C_3)\le
C_1x^{2\alpha-2}(-\tau^{\alpha-1/2}+C_3)\le0
\end{align*}
for small enough $\tau$.

From 1) and 2) it follows that $Lv\le0$ in some $Q_{\varepsilon,\delta}$ which finishes the proof that $u_x(0,t)\ge C(1-t)^{\alpha-1/2}$.

Also, answering your question from another post, it follows from the inequality $u\ge K_1v$ in $Q_{\varepsilon,\delta}$ that $u(x,1)\ge C x^{2\alpha}$.