Symbolic Integration gives different results by changing the constants. Is it a bug?

Both values seem to be correct, depending on the parameters; however, the dependence on the parameters is not correctly expressed in the ConditionalExpression result. It may be that the OP has in mind a real integral, but Mathematica, by default, treats the parameters as complex numbers. By a standard application of the Residue Theorem, an integral of the form $(1/2\pi)\int_0^{2\pi} R(\cos t,\sin t)\,dt$, where $R(x,y)$ is a rational function, is equal to the sum of the residues of $$ {1\over z}\cdot R\left({z+z^-1 \over 2},{z-z^-1 \over 2i}\right) $$ over the poles in $|z|<1$. The OP's integral is of this form ($2\theta=t$): $$ \int_0^\pi \frac{ \Delta p^2 (1-F) \sin^2(\theta -\theta_r+ \Omega\tau )}{\pi \left(1-\Delta p^2 \cos ^2(\theta -\theta_r+ \Omega\tau )\right)} \; d\theta = \int_0^{2\pi} \frac{ \Delta p^2 (1-F) \sin^2(t/2-t_0)}{2\pi \left(1-\Delta p^2 \cos ^2(t/2-t_0)\right)} \;dt $$ $$ \qquad= {1\over2\pi}\int_0^{2\pi} \frac{\Delta p^2 (1-F) (\sin t \sin 2 {t_0}+\cos t \cos 2 {t_0}-1)}{\left({\Delta p}^2+{\Delta p}^2 \sin t \sin 2 {t_0}+{\Delta p}^2 \cos t \cos 2 {t_0}-2\right)} \;dt $$

Here is are plots of the numerical integration for an imaginary value of $t_0$ (code below):

Fig. 1. Plots of the integral computed with NIntegrate[] for $\Delta p=x+i\,y$, $F=0$, $t_0=0.1i$; the middle plot shows that the integral equals $1-F = 1$ outside a figure-eight region (where all the poles are inside the unit disk).

Here is an implementation of the above idea:

op = 1/π (Δp^2 Sin[Ω τ + θ - θr]^2 (1 - F))/(1 - Δp^2 Cos[Ω τ + θ - θr]^2) /. 
    {θr -> t0 + Ω τ, θ -> t/2};
integrand = op/2 // TrigReduce // TrigExpand // 
   Simplify[#,                 (* Complexity function unwanted trig forms *)
     ComplexityFunction -> (LeafCount[#] + 
         1000 Count[#, 
           Sin[1/2 (t - 2 t0)] | Sin[t/2 - t0] | Cos[t - 2 t0], 
           Infinity] &)] &;

intz = Block[{F = 0},
     integrand
     ] /. {Sin[t] -> (z - 1/z)/(2 I), Cos[t] -> (z + 1/z)/2} // Simplify;
poles = Solve[z*Denominator[intz] == 0, z] // Simplify;
residues = Residue[1/z*intz, {z, z /. #}] & /@ poles // FullSimplify;

(* the value of the integral *)
val = 2 Pi*residues.UnitStep[1 - Abs[z /. poles]] // PiecewiseExpand // FullSimplify

The third condition cannot happen, but Mathematica does not figure that out. (The two distinct absolute values are reciprocals of each other, and the greater one has a minimum value of 1. This implies the first two pieces cover all cases. But that's difficult to compute.)

Fig. 2. Plots of the symbolic value val of the integral for $\Delta p=x+i\,y$, $F=0$, $t_0=0.1i$; the middle plot show that the integral equals $1-F = 1$ outside a figure-eight region (where all the poles are inside the unit disk).

Remark on the computational issue. One reason for the different results that the OP experiences is probably the increase in complexity of using a product of constants Ω τ over a single constant a. I think this is worth emphasizing: Mathematica will try to reduce the conditions and perhaps tries to break down the conditions into terms involving the parts of Ω and τ separately. This might lead to different results than with a simpler a. (It does not, however, explain the incorrectness in the result.) Sometimes, with symbolics it pays to simplify the parameters, such replacing Ω τ by a, or indeed both terms Ω τ - θr by t0. They can then be substituted back at the end of the computation.

---

Code for figure 1:

Block[{Δp = x + I y, t0 = 0 + t1 I, t1 = 0.1, F = 0},
  iplotAbs = Plot3D[
     Abs@NIntegrate[
       1/π (Δp^2 Sin[t - t0]^2 (1 - F))/(1 - Δp^2 Cos[t - t0]^2),
       {t, 0, π}, Method -> "Trapezoidal", PrecisionGoal -> 4, 
       AccuracyGoal -> 4, MaxRecursion -> 20],
     {x, -1.04/t1, 1.04/t1}, {y, -1.04/t1, 1.04/t1}, 
     MaxRecursion -> 3, PlotLabel -> "Modulus of integral", 
     MeshFunctions -> {#3 &}, AxesLabel -> {x, y}
     ];
  ] // AbsoluteTiming
(*  {141.737, Null}  *)

Block[{Δp = x + I y, t0 = 0 + t1 I, t1 = 0.1, F = 0},
  iplotArg = Plot3D[
     Arg@NIntegrate[
       1/π (Δp^2 Sin[t - t0]^2 (1 - F))/(1 - Δp^2 Cos[t - t0]^2),
       {t, 0, π}, Method -> "Trapezoidal", PrecisionGoal -> 4, 
       AccuracyGoal -> 4, MaxRecursion -> 20],
     {x, -1.04/t1, 1.04/t1}, {y, -1.04/t1, 1.04/t1}, 
     MaxRecursion -> 3, AxesLabel -> {x, y}, MeshFunctions -> {#3 &}, 
     PlotRange -> All, PlotLabel -> "Argument of integral"
     ];
  ] // AbsoluteTiming
(*  {145.414, Null}  *)

GraphicsRow[{iplotAbs, 
  Show[iplotAbs, PlotRange -> {0.8, 1.2}, ViewPoint -> {0, 3.5, 0.7}], 
  iplotArg}, ImageSize -> 600]

Code for Figure 2:

Block[{Δp = x + I y, t0 = 0 + t1 I, t1 = 0.1, F = 0},
 GraphicsRow[
  Plot3D[
     #[val],
     {x, -1.04/t1, 1.04/t1}, {y, -1.04/t1, 1.04/t1}, 
     MaxRecursion -> 3, MeshFunctions -> {#3 &}, ImageSize -> 300, 
     AxesLabel -> {x, y}, PlotLabel -> #["val"], 
     Exclusions -> {Automatic, {x == 0, # === Arg}}
     ] & /@
   {Abs, Arg}, ImageSize -> 600
  ]
 ]

With suitable Assumptions, the two formulations give equivalent results:

Integrate[1/π (Δp^2 Sin[Ω τ + θ - θr]^2 (1 - F))/(1 - Δp^2 Cos[Ω τ + θ - θr]^2), 
    {θ, 0, π}, Assumptions -> -1 < Δp < 1 && 0 < θr < π && 0 < Ω τ < π]
(* ConditionalExpression[(-1 + F) (-1 + 2 Sqrt[1 - Δp^2]), π + 2 θr < 2 τ Ω] *)

Integrate[1/π (Δp^2 Sin[a + θ - θr]^2 (1 - F))/(1 - Δp^2 Cos[a + θ - θr] 2), 
    {θ, 0, π}, Assumptions -> -1 < Δp < 1 && 0 < θr < π && 0 < a < π]    
(* ConditionalExpression[(-1 + F) (-1 + 2 Sqrt[1 - Δp^2]), 2 a > π + 2 θr] *)

Both now are wrong! The correct solution is

Integrate[1/π (Δp^2 Sin[θ]^2 (1 - F))/(1 - Δp^2 Cos[θ]^2), {θ, 0, π}, 
    Assumptions -> -1 < Δp < 1]
(* (-1 + F) (-1 + Sqrt[1 - Δp^2]) *)

Setting Ω τ - θr to zero is valid, because the integrand is periodic with period Pi. Incidentally, setting only Ω τ to zero also gives the correct result, although with an extraneous condition.

Integrate[1/π (Δp^2 Sin[θ - θr]^2 (1 - F))/(1 - Δp^2 Cos[θ - θr]^2), {θ, 0, π}, 
    Assumptions -> -1 < Δp < 1 && 0 < θr < π]
(* ConditionalExpression[(-1 + F) (-1 + Sqrt[1 - Δp^2]), 2 θr < π] *)

Clearly, the strange behavior exhibited here and in the question represents one or more bugs.

These results were obtained with

$Version
(* "10.4.1 for Microsoft Windows (64-bit) (April 11, 2016)" *)

and

(* "11.2.0 for Microsoft Windows (64-bit) (September 11, 2017)" *)

Incidentally, the newer version of Mathematica does not reproduce the ConditionalExpression in the question, instead giving,

(* ConditionalExpression[-I (-1 + F) (-I + Sqrt[-1 + Δp^2] + 2 Sqrt[-1 + Δp^2]
   Floor[Arg[-1 + Δp^2]/(4 π)]), 
   Im[a] == Im[θr] && -(π/2) + Re[a] <= Re[θr] <= π/2 + Re[a]] *)

Addendum

Identical behavior persists for

(* "11.3.0 for Microsoft Windows (64-bit) (March 7, 2018)" *)

except that it fails in a different way to reproduce the ConditionalExpression in the question, instead giving,

ConditionalExpression[1 - F, C[1] ∈ Integers && 
    Im[a] == Im[θr] && -(π/2) <= π C[1] - Re[a] + Re[θr] <= π/2]