When $x\to0$, the LHS is equivalent to $4^{1-a}\pi x$ and the RHS is equivalent to $x^{a-1}$. If $a\gt2$, $x^{a-1}\ll x$ hence the inequality cannot be true on a neighborhood of $0$.
If $a=2$, the inequality holds if and only if $u(x)\geqslant0$ for every $x$ in $(0,1)$, where $ u(x)=4x(1-x)-\sin(\pi x). $ Since $u(1-x)=u(x)$, one can, and we will, assume that $x$ is in $(0,1/2)$. Since u''(x)=-8+\pi^2\sin(\pi x) increases from u''(0)=-8 to u''(1/2)=\pi^2-8\gt0, u'' is first negative then positive on the interval $(0,1/2)$, hence u' decreases then increases. Since u'(x)=4(1-2x)-\pi\cos(\pi x), u'(0)=4-\pi\gt0 and u'(1/2)=0. Thus u' is positive then negative on the interval $(0,1/2)$. Since $u(0)=u(1/2)=0$, $u\geqslant0$ on the interval $(0,1/2)$. This concludes the proof for $a=2$.
For a given $x$ in $(0,1)$, the ratio of the RHS by the LHS is $(4x(1-x))^{a-1}$ times a factor independent on $a$. This is a nonincreasing function of $a$ because $0\leqslant4x(1-x)\leqslant1$. Hence the inequality being true for $a$ implies it is true at every smaller value of $a$.
Finally the inequality holds for every $a\leqslant2$ and it fails for every $a\gt2$.