This is a most confusing question. I withdrew my answer to the first version because I felt dissatisfied with the extrapolation to an 'ellipse' which would be a segment. And I think it is wrong and you have no right to apply Kepler's laws to such a trajectory. Mathematically, an ellipse turns into a parabola when eccentricity -->1 or it can degenerate into a pair of straight lines when you write the general conic section equation and impose certain conditions on the coefficients. But a segment, I don't see how. Physically, a trajectory through the center of force is completely different from a trajectory which has angular momentum.
Kirchway's equality KE = -PE/2 holds only because he modifies the assignement and takes 2Pi au/y velocity, which allows him to apply results valid for the earth. But of course KE=-PE/2 does not hold generally for a trajectory which ends in a finite time (the demonstration of the virial theorem does not hold) It does not hold if you take 1 au/y as initial velocity of your degenerate trajectory.
Flamechop and Magnetulsar are completely wrong since even if you had postulated a 2Pi au/y velocity, it would still be short by a factor sqrt(2) from the escape velocity. (and I assume 8th grade astronomy class is a joke)
So your answer can be found by integrating 1/sqrt[2(GM/r+C)] with C = v_0^2/2 - GM/Re first from Re to Rmax (with Rmax from C = -GM/Rmax) and then from Rmax to 0, but I don't see any way to derive it using Kepler's laws. Even if you did use them for the segment 0-Rmax seen as an ellipse of semi major axis Rmax/2, (which is not warranted in my opinion but it might be that the integral gives you the right relation for the 'half year'-you'll only know it by integrating !) you would still have to calculate the time from Re up to Rmax using integration.
Edited: I have just skimmed through dear old Lev Davidovitch and in (very) short: conservation of angular momentum (L) is equivalent to the area law which by integration yields 2m_eS = Te L with S the orbit area. Knowing it is an ellipse and having related the semi axes to G, M, L, E and m_e by using the conservation laws to solve for theta(r), you can eliminate both S and L (and m_e) and get Kepler's third law. Of course this one no longer depends on finite surface or non zero angular momentum, but you simply cannot derive it without using them.
To Sum Up: if you start from Newton, which is arguably the basis, there is no way to deduce Kepler's law for a zero angular momentum trajectory. (Besides, they are absolutely meaningless for such a trajectory) You have one and only one equation which is d^2r/dt^2 = -GM/r^2 which gives you a first (energy) integral, 1/2dr/dt-GM/r = C and by integrating that one can answer your question, but there is no way to short cut calculus.
Re: kirchway, of course I agree with KE = -PE/2 (for any periodical orbit actually) so you can use earth data to calculate initial KE in terms of Rearth by scaling the speed and then find Rmax.
My contention is that you have no right to use T^2 =kR^3 for the degenerate orbits going through the center because to demonstrate it, you need T to be a period (not a countdown time to an end) and R to be the SMA of a finite area ellipse.
This is a question of physics, not of math. Now since there is no apparent discontinuity in the formula, you can bet it is usable for the limiting case also. However you can only be sure by explicit integration (and of course, it works)
It remains that
a/ even with this somewhat dubious 'Kepler's law', you still have to perform an integration for the Re-->Rmax part (Done in your complicated formula)
b/ your final answer must be wrong..see below
LiD's remark about relativity not being important is valid only thanks to the finite solar size. Otherwise the speed grows without limit to compensate for PE --> -infinity and keep the total energy constant and finite. Then at some point one can no longer claim v/c << 1 Because of that, LiD's time for the pointlike sun is certainly wrong and so is yours if your sun is pointlike
Actually there is an almost consistent way to consider the degenerate trajectory as the limit of an ellipse by using the angular momentum L as a parameter -->0 Keeping the energy fixed, the eccentricity-->1 (1-e^2 ~ L^2) the major axis is fixed, the minor axis-->0 (b ~ L) and the point of closest approach ->0 as L^2
The inconsistency lies in that this point is reached for theta=pi, that is, the 'other side' of the center with respect to what happens with the L=0 trajectory. Clearly, there can't exist a consistent way of treating angles when there are none in the limit.
The approximate time up calculation of AI P is beautiful but evidently wouldn't work with v=2Pi au/y