How strokers affect how a cam "acts"

If you are looking for off-the-line snap;
I think you are gonna want a bit more gear and/or a bit more stall. And here's why; From personal experience with cars, I have found that to be really snappy off the line,the engine needs to be loaded with less than 1.4 pounds per ftlb of torque at the ground.
How did I get to that? Well, check out this hot 360 dyno graph;
http://www.hughesengines.com/Upload/TechArticles/land_dyno.jpg
This engine will take off very well in a 3650pound car with about 8.7/1 starter gear and 27 inch tires, and a 2800TC. So looking at the graph and extrapolating the torque curve down to 2800, I get about 340 ftlbs there. Ok, now multiply that by the reduction ratios in the tranny and rear, and I get 340 x 2.45 x 3.55=2957.. Now correct that for the 27s {12/(1/2x27)}and I get ; .889x2957=2628 to the ground. Finally 3650/2628=1.39 and this is pounds of vehicle per ftlb output, at the crank. This makes a very snappy take off.
I did the math for a sluggish combo, namely a teener with a 2.45x 2.73 gear and 24 inch tires,and 3350pounds and it comes to about 1.8 pounds per ftlb.
Your combo,guessing at 340ftlbs@2400, with 2.45x4.10s and corrected for 32s would math out to 1.95 pounds per ftlb.( I guessed at the engine output). So your combo might be a little slower than sluggish(1.8);if I guessed right at the 340@2400. Remember this is the start-up gear only.But this start up gear will take the truck to about 55mph at the top of first gear.
To get up to teener status (1.8) would require 5000/1.8=2778 output, and that maths to 369 crank ftlbs.I think a bit of TC will get you there.
To be really snappy,your combo will want; (5000/1.4=),3571ftlbs to the ground, just to equal that above 360, in its 3650 chassis. So with 32s and 4.10s, and a 2.45low, your engine will need to put out 474 ftlbs at your chosen stall of 2400. I don't think that is doable with either cam.So I'm thinking you need more gear,AND more TC.
This is all just math.
If you put 4.88s in it, the requirement would come down to 398 ftlbs, and I think that is doable, with just a bit more TC.

Again this is all just math and guess work on my part,and I'm only trying to be helpful.If you can find a dyno graph for an engine like your proposed one, you can do your own math.
Here is a formula I just worked out;
crank torque required= chassis weight/Q-factor x (1/ R1R2x cF)
Where;Q-factor is how snappy you want it be;1.4very to 1.8not-so-much, R1R2 are the torque multipliers, and cF is the tire correction of (12/.5x the tire diameter in inches). Your correction is 12/.5x32=.75... Your torque multipliers are 2.45 x 4.10=10.045. Your weight is 5000.Your Q-factor is between 1.4 and 1.8
So here's an example;crank torque=5000/1.6 x [1/(2.45x4.1x.75)]= 415
So then just look on the graph, find 415 ftlbs, and read the rpm. That is the stall you will need; in this example.
Like I said, just trying to be helpful.