EDITED. The major obstacle when calculating the integral
$$ I
= \int_{1}^{\infty} (u^{\alpha} - (u-1)^{\alpha})^{2} \, du
= \int_{0}^{1} x^{-2\alpha-2}(1 - (1-x)^{\alpha})^{2} \, dx $$
is that, within the range $|\alpha| < \frac{1}{2}$, the term $x^{-2\alpha-2}$ has singularity at $x = 0$ which gives rise to infinity when it is not cancelled out appropriately. In order words, splitting the integral into three parts gives
\begin{align*}
I_{1} &= \int_{0}^{1} x^{-2\alpha-2} \, dx = \infty, \\
I_{2} &= -2 \int_{0}^{1} x^{-2\alpha-2}(1 - x)^{\alpha} \, dx = -\infty, \\
I_{2} &= \int_{0}^{1} x^{-2\alpha-2}(1 - x)^{2\alpha} \, dx = \infty.
\end{align*}
One way of avoiding this catastrophe is to use the principle of analytic continuation. To utilize this technique, let us introduce auxiliary integral
$$ I(s)
= \int_{1}^{\infty} u^{-s}(u^{\alpha} - (u-1)^{\alpha})^{2} \, du
= \int_{0}^{1} x^{s-2\alpha-2}(1 - (1-x)^{\alpha})^{2} \, dx. $$
For any given $|\alpha| < \frac{1}{2}$, it is not hard to check that $I(s)$ converges and defines an analytic function for $\Re(s) > -2\alpha-1$. Indeed, from the mean value theorem we find that $u^{\alpha} - (u-1)^{\alpha} = \Theta(u^{\alpha-1})$ as $u \to \infty$. Consequently, the integrand is asymptotically
$$ u^{-s}(u^{\alpha} - (u-1)^{\alpha})^{2} = \Theta(u^{-\Re(s)+2\alpha-2}) $$
and hence it becomes integrable.
Moreover, for $\Re(s) > 1+2\alpha$, we have $\Re(s-2\alpha-2) > -1$ and hence we become able to evaluate $I(s)$ by expanding the square and integrating each resulting term as follows:
\begin{align*}
I(s)
&= \int_{0}^{1} x^{s-2\alpha-2} \, dx - 2 \int_{0}^{1} x^{s-2\alpha-2}(1 - x)^{\alpha} \, dx + \int_{0}^{1} x^{s-2\alpha-2}(1 - x)^{2\alpha} \, dx \\
&= \beta(s-2\alpha-1, 1) - 2\beta(s-2\alpha-1, 1+\alpha) + \beta(s-2\alpha-1, 1+2\alpha) \\
&= \Gamma(s-2\alpha-1) \left( \frac{1}{\Gamma(s-2\alpha)} - 2\frac{\Gamma(1+\alpha)}{\Gamma(s-\alpha)} + \frac{\Gamma(1+2\alpha)}{\Gamma(s)} \right). \tag{1}
\end{align*}
Now notice that the right-hand side defines a meromorphic function on all of $\Bbb{C}$. So by the principle of analytic continuation, (1) holds for all $\Re(s) > -2\alpha-1$ away from poles of $\Gamma(s-2\alpha-1)$. Now taking limit as as $s \to 0$, we obtain
\begin{align*}
I
&= \Gamma(-2\alpha-1) \left( \frac{1}{\Gamma(-2\alpha)} - 2\frac{\Gamma(1+\alpha)}{\Gamma(-\alpha)} \right) \\
&= -\frac{1}{2\alpha+1} - 2\beta(-2\alpha-1, 1+\alpha),
\end{align*}
This coincides with Lucian's answer, and also coincides with numerical tests with Mathematica.
OLD ANSWER. It seems that the exact closed form for the integral seems hard to obtain. Instead, some approximations are possible. For example, I obtained
$$ \frac{2\alpha^{2}}{1-4\alpha^{2}} \leq I \leq \frac{2\alpha^{2}}{(1-4\alpha^{2})(\alpha + \frac{1}{2})}. $$
In particular, we have $I = \Theta(\alpha^{2})$ near $\alpha =0$ and $I ~ \sim (4-8\alpha)^{-1}$ as $\alpha \to \frac{1}{2}^{-}$.
