So are you saying that we could not compensate for the torque ?

Net torque is required. If you like you can think of the reduction of angular momentum as reduction of the Kinetic Energy you have with respect to referece frame of the fixed stars while sitting in your chair, to that of an object in a highly elliptic orbit at its 1AU apogee. I.e. give the nuclear waste rocket an orbit with apogee of 1 AU and perigee equal to the sun's radius, not the nearly circular 1AU orbit is has when sitting on the launch pad. If it is easy to do, I'll compute how much work is required and edit this post.

By Edit: Must essentially stop all its forward orbit speed:

Following is my “stream of thought” approach to problem of computing how much work must be done to send each unit of mass into the sun. Answer, I think correct, is you must remove essentially 100% of the kinetic energy the Nuclear Waste Rocket had just sitting on the launch pad.

Taking Earth to be in a 1 AU circular orbit the KE = 0.5|PE| and in some strange, but convenient “Billy T scale (BT units) is: 0.5M[2pi(1AU)/365]^2 which I then “normalize” per BT unit of mass to have the PE = -2 and earth's KE =1.

The radius of the sun, is: 0.00464913034 AU (Ain't Google wonderful) but I'll call it 0.00465 and note gravitational potential is inverse linear. Thus the PE falls from -2 down to -2 / 0.00465 = -430. (in TBs units, normalized of course) from what the slowed down nuclear waste rocket, NWR, had at 1 AU apogee and it gains 238 units of KE in the “fall” down to perigee.

Now since PE is always defined wrt some arbitary zero point, and I have the KE at perigee known to be increased from the tiny value it had at 1AU apogee (after slowed down form Earth's orbit speed) which I'll guess was 0.3, so I can redefine the PE at 0.00465Au to be zero (and must add 430 units of energy to all previous values given.)

Thus the PE at apogee of the elliptical orbit NWR's unit of mass with new zero at 1 AU is -2 + 430 = 428 and by guess of 0.3 for the KE there, the total energy there is 428.3, but it is constant (until drag in the solar corona takes effect). Thus the work done, W, per unit mass to slow each unit of mass of the NWR down from Earth's orbital speed, to the elliptical orbit apogee's speed is W = 428.3 – 428 = 0.3, my guess returning. I. e. 70% of the KE it had on Earth must be removed by work slowing it down, but this is only a guess.

To do it more correctly we need to calculate the KE the “slowed down” NWR had at 1AU. This can be done by noting the elliptical orbit sweeps out equal areas each micro second everywhere in it orbit.It had KE = ~428 at perigee. I. e. the travel distance in a micro second at 0.00465AU is 430 greater (by faster speed) than at 1AU and the KE at perigee is 430^2 = 184,900 times greater. Thus instead of having KE of 1 when on the launch pad, The NWR must have essentially all it KE removed by the work done! Not my guess of only 70%