Answer
Yes
Work Step by Step
Since the collision is elastic, the kinetic energy of the system will be conserved and the objects will fly apart separately after they collide. Therefore, the cue ball will not travel in the same direction as the red ball and fall into the same hole.
However, it is possible for it to fall in the right hole.
Assume the red ball makes an angle of $\theta$ and the white ball makes an angle of $\alpha$ with the vertical.
$P_{ix}=P_{fx}$
$P_{iy}=P_{fy}$
$mv=mv'_1cos(\theta)+mv'_2cos(\alpha)$
$0=mv'sin(\theta)-mv'_2sin(\alpha)$
$v^2=v'^2_1cos^2(\theta)+2v'_1v'_2cos(\theta)cos(\alpha)+v'^2_2cos^2(\alpha)$
$0=v'^2_1sin^2(\theta)-2v'_1v'_2sin(\theta)sin(\alpha)+v'^2_2sin^2(\alpha)$
$v^2=v'^2_1+v'^2_2+2v'_1v'_2cos(\theta+\alpha)$
$E_{iKx}=E_{fKx}$
$mv^2=mv'^2_1cos^2(\theta)+mv'^2_2cos^2(\alpha)$
$E_{iKy}=E_{fKy}$
$0=mv'^2_1sin^2(\theta)+mv'^2_2sin^2(\alpha)$
$v^2=v'^2_1+v'^2_2$
$v'^2_1+v'^2_2+2v'_1v'_2cos(\theta+\alpha)=v'^2_1+v'^2_2$
$2v'_1v'_2cos(\theta+\alpha)=0$
$cos(\theta+\alpha)=0$
$\theta+\alpha=90^o$
Using simple trigonometry (toa), we can find that $\theta=30^o$ and $\alpha=60^o$
Then, the ball will enter the right hole.