If you're talking statics (i.e. the "weight in his hand" as you say, but without any motion), then it should be roughly:
W = (1/2)*F*cos(theta)
where theta is the angle between the floor and the plane of the tyre and F is the weight of the tyre (1000lbs). The factor of 1/2 comes from the leverage (the location of his hands is twice as far from the pivot as is the centre of mass).
So, the "weight in his hands" will be roughly 500lbs at zero degrees, and 250lbs at 60 degrees. At 45 degrees it would be roughly 353lbs.
[Edit: You asked for the proportion, so that will be 0.5*cos(theta). Also note I'm using "weight" here in the colloquial sense - it should really be the "mass" of the tyre, and the "weight" in his hands would be multiplied by g to obtain the force in Newtons.]
To compute the exact force he is exerting you need a somewhat more complex calculation, which takes into account the moment of inertia, the rate at which he flips the tyre, and the angle that he directs the force (which won't be constantly vertical). Either way, the force exerted will be greater than in the static case.
