If ice is "all around the sun" I fail to see how it can be moving at a velocity of 1000 m/s inwards. The mass of the sun is $2\cdot 10^{30}\mathrm{\;kg}$ and the radius $7\cdot 10^{8}\mathrm{\;m}$.
The thickness of a shell of ice with that inner radius and mass would be (assuming the usual density of ice of about 0.9x that of liquid water) approximately $10^8 \mathrm{\;m}$. That is probably sufficiently thick to withstand the gravitational attraction of the sun; certainly it will stop most solar radiation.
To take one gram of ice from absolute zero to melting takes roughly 273*4.2+334=1500 J. The sun puts out about $4\cdot 10^{26}$ W - assuming that the ice would absorb all that heat, it would take $8\cdot 10^9$ seconds to melt all the ice - a bit more than 200 years.
All that time the sun would be happily continuing to produce power - but I expect all life on earth would have ceased by the time it starts to shine again.
One obvious question - would the spherical "ice shell" be stable undertake the tremendous gravitational stress? Would it melt under the pressure? Would the pressure of the steam generated blow the water / ice outwards? It would be interesting to analyze those questions further. I suspect the over all conclusion - that a 100,000 km thick layer of water will "turn out the lights on Earth" will be unaltered by these details - because that water will still be between the Earth and the Sun, regardless of the distance and the phase.
update - a few additional thoughts.
First - the crushing strength of ice is quite low: no more than about 1000 psi (7 MPa) according to this USGS report. That is obviously many orders of magnitude smaller than the pressure on the inside of the 100,000 km thick ice shell. The average distance of the shell (mid point) is $7.5\cdot 10^8 \mathrm{\;m}$ from the center of the Sun, and will therefore experience a gravitational acceleration of
$$a = \frac{GM}{R^2} = \frac{6.7\cdot 10^{-11} \cdot 2\cdot 10^{30}}{(7.5\cdot 10^8)^2} = 240 \mathrm{\;m/s^2}$$
Thus the pressure on the inner surface is roughly
$$P = \rho a t = 0.9E3 \cdot 240 \cdot 1E8 = 2.5 \cdot 10^{13} Pa = 22 TPa$$
An obvious question to ask: what happens to ice at that pressure? The phase diagram I could find (at this location) "only" goes up to 1 TPa, but it suggests that "really cold" ice does in fact remain solid at these pressures (unlike slightly warmer ice like we normally encounter, this would be "phase XI hexagonal" ice).

The next interesting question is that of steam formation. If we did drop a certain volume of ice into the sun (inside the closed ice shell), what happens to the pressure? Presumably the pressure would increase somewhat, but it really isn't relevant - because again, at the pressure you would have to generate to support the ice shell, the density of the water would have to be very high - in fact, it would no longer be a gas, but a solid (or at least with comparable density to a solid - we would be in a part of the phase diagram that is not given).
Finally, the question of the potential energy of the ice - and the impact of the release of this energy on the over all equation. For the purpose of this calculation, we can't just assume that things fall to the center of the sun - even photons that are generated at the center of the sun take a long time to diffuse to the surface, so we can assume the same is true for water. Let's assume therefore that the water simply falls to the surface. While the inside of the shell only falls 1000 km, on average the ice would fall 50,000 km. The force of gravity can be considered (to first order) constant over this distance, so the work done on 1 kg of ice would be
$$W = F\cdot d = 240 N \cdot 5\cdot 10^{7} m = 12 GJ$$
The ice that fell from the inside of the shell (the first ice that melts) has less energy, namely
$$W = 240 N \cdot 10^6 m = 240 MJ$$
and I am for now ignoring the claim that the ice is moving at 1000 km/s (from the original question) as that would mean 1 kg of ice had a kinetic energy of $\frac12 m v^2 = 500 GJ$.
Whichever way you look at it, that is a very considerable amount of energy. It suggests that as the ice starts melting from the inside of the shell out, the water slamming into the surface of the sun will actually heat the sun up, speeding up the melting of the rest of the ice. The whole process will therefore take much less time than I initially estimated - there will be a runaway reaction.
Just to calibrate us - all that ice slamming into the sun adds about 12 GJ/kg * 2E30 kg = 2.4E40 J to the sun. If none of that energy was transferred to the sun, it would lead to a temperature rise of the water of about 3 million degrees. Just from the potential energy (not the initial kinetic energy). That is much hotter than the sun - so there would be a runaway melt reaction.
So it seems that after a brief time when the sun is dark (much less than 200 years), it would shine very, very brightly? Still seems like an uncomfortable solar system.
UPDATE 2
One more thought. If the ice was a little bit less dense, so that structurally it can all fall to the surface of the sun, the 100,000 km thick layer of ice would (at an initial velocity of 1000 km/s) take just 100 seconds to fall into the sun. On average, each bit of ice would fall just 1000 km, and most of the energy dissipated would be the inital kinetic energy (500 GJ/kg - much more than the 240 MJ/kg gravitational energy).
This would briefly heat the surface of the sun to a temperature of more than 100 million degrees - hotter than the core of the sun. So in that case the sun might blink briefly (while the ice is still acting as a shield) - but very quickly, it would all be over for the earth. Of course at that temperature all kinds of fusion reactions would take place - and an immense amount of heat would radiate from the surface of the sun.
It reminds me of Tom Lehrer's song - "We will all go together when we go"
There will be no more misery
when the world is our rotisserie
yes we will all fry together when we fry.