Take the periodicity of the diffequation into account. Since t only occurs in Sin or Cos with minimal 10 t, you have peroidicity of 2 Pi/10. NDSolve for high t would be very unreliable; integrate within lower period (But not from the beginning t==0 because of disturbing effects due to initial conditions). Rationalize parameters.

sbar = 420;
theta = 1;
h = 1/2;
w2 = 10;
nu = 7/10;
al[t_] = sbar Sin[2 w2 t];
rho[t_] = 1 + nu Sin[w2 t]; 
r[x_, t_] = -theta (1 - x) D[rho[t], t] // Simplify;
r1[x_, t_] = -theta x (1 - x) al[t] rho[t] (1 - 2 x + h (4 x - 3)) // 
 Simplify;

(eqs = Subtract @@@ {D[v[x, t], 
       t] == -al[t] x (1 - x) D[(x + h (1 - 2 x)) v[x, t], 
         x] + (x (1 - x)/(2 rho[t])) D[v[x, t], x, x] + r[x, t] + 
       r1[x, t], v[x, 0] == 0, v[0, t] == 0, v[1, t] == 0} // 
   Together // Numerator // Expand // Simplify) // TableForm

vsol = v /. 
  First@NDSolve[Thread[eqs == 0], v, {x, 0, 1}, {t, 0, 3}, 
MaxSteps -> 10^5, MaxStepSize -> {.0002, .04}, 
StartingStepSize -> {.0001, .01}]

Periodiity in t for all x shown in graphic. Integrating from t=2 Pi/w2 to t=4 Pi/w2 should yield the same as your given range.

Manipulate[
  Plot[{vsol[x, t], vsol[x, t + 2 Pi/w2], vsol[x, t + 4   Pi/w2]}, {t, 
  2 Pi/w2, 4 Pi/w2}, PlotPoints -> 100], {{x, 1/2}, 0, 1}]

Plot3D[vsol[x, t], {x, 0, 1}, {t, 2 Pi/w2, 4 Pi/w2}, PlotRange -> All,
 PlotPoints -> 50]

nint1 = NIntegrate[2 vsol[x, t], {x, 0, 1}, {t, 2 Pi/w2, 4 Pi/w2}]

(*   -0.00674974   *)

(*   Error message: ....NIntegrate obtained -0.00674974 and 6.36785*10^-8 for the integral \
and error estimates   *)

If you are not satisfied with the error of 6 10^-8, you have to use higher WorkingPrecision. But this takes very long and may crash to not enough memory.

Integrate[
 2 (1 - x) rho[t], {x, 0, 1}, {t, 42 Pi/w2, 44 Pi/w2}]/(2 Pi/w2) + 
nint1/(2 Pi/w2)

(*   0.989257   *)