"So, if falling from Jupiter, one is only going to get a velocity less than 10% of Sun's escape velocity [around 60 km/sec]"

by the time it gets to to mercury orbital distance.

So how much ice melts with 63.08 Million Watts/m2.

So, 335000 Joules/kg of 0 C ice to turn to water.
http://van.physics.uiuc.edu/qa/listing.php?id=1573

So, if all solar radiation energy is absorbed as heat one gets
63,080,000 joules per sq meter per sec.
So if you had a meter sq of ice which was 10 cm thick, it would require 33,500,000 joules of heat to melt.
And 2000 Joules/kg to increase temperature by each C, so
200,000 times 100 to reach 100 C, so additional 20 million Joules.
To boil 1 kg of water requires 2.26 million joules [or in this case minimum amount of heat needed to expel as propellent]. So the meter sq by 10 cm require 226 million joules to boil.
33,500,000 + 20,000,000 + 226,000,000=
279.5 million Joules per 10 cm and
2.795 billion Joules per meter depth of ice which if 100% of the solar energy absorbs turns it to steam in 44.3 seconds. And 10 meter thick in 7.38 minutes.
Since one couldn't adsorb 100% it probably be about
10 meters per 10 minutes.

So the 63.08 Million Watts/m2 is at the Sun surface, you wouldn't want to fly to the sun's surface nor thru the Sun's atmosphere: "The photosphere has a particle density of about 10^23 m−3 (this is about 1% of the particle density of Earth's atmosphere at sea level)"
And "The photosphere is actually tens to hundreds of kilometers thick"
http://en.wikipedia.org/wiki/Sun

Probably the closest possible is 1000 to million kilometers above the photosphere:
"The low corona, which is very near the surface of the Sun, has a particle density of 10^14–10^16 m−3. (Earth's atmosphere near sea level has a particle density of about 2×10^25 m−3.)"
But tens of million to a billionth of Earth sea level atmosphere could still be too dense if traveling at hundreds of km/sec.

So, if the closest distance is about 1/2 of the Sun radius- around 350,000 km above the surface, one will have around 32 million watts per sq meters and escape velocity of around 490 km/sec

And if closest distance is about 1/4 radius- around 173,000 km, it's a escape velocity 536 km/sec and around
40 million watts per sq meter.

So, if you need to go thru a solar atmosphere of less than 1 x 10^14 per m-3 you probably could only get about 90 to 95% of sun escape velocity and will have a peak of around 50 million watts.

So, a question could be what would the duration that one would have 1/10th or more [6 million watts sq meter] of the energy that turns 1 meter of ice into steam in 44.3 seconds.

So more than 1/10th it would start and end at a distance which had 10 times the surface area as the Sun's surface, so this about 3 times the radius or around 2 million km from the Sun.
Escape velocity at this distance is about 346 km/sec.

So a 1/2 circle 3 million km in diameter is about 4.7 million km. So 1/2 this is about 2 million km to reach a 1/2 or 1/4 Sun radius distance above surface. So at 350 km/sec it takes 5714 seconds to go 2 million km. And at
400 km/sec it take 5000 seconds.
So you pick 30% of "44.3 seconds rate" it take 120 second per meter, so you need about 50 meter ice.
If you leave at 530 km, it about 3800 seconds, requiring 32 meters.

So you need about 100 meters thick of ice to function as heat shield and source rocket propellent.
The area could be about 4 sq km, so this mass would be about 400 million tonnes. Add about 10% of this mass for tankage, plumbing, pumps, rockets, and 20% as payload,
so less than 450 million tonnes in total mass.

So, this might have about 20 km/sec of rocket delta-v, but adding this delta-v at the near the bottom of Sun gravity well will make significant more efficient in term adding to velocity to escape, you could also possible add 10 km/sec from gravity assist from planet, and one could get some percentage the galactic orbital velocity. So something in total as much as 200 km/sec escape velocity from Sol.