This might not be as rigorous as desired. Hopefully, it's correct.

Suppose $\Gamma_{\psi}$ is not fully symmetric. There is some symmetry element $x$ whereby $x$ is antisymmetric in this representation.

Now, consider this operation. You could perform the change of variable $x\tau \rightarrow \tau'$ for the integral, and I believe this requires a coordinate change on $\psi$ as well.

$$\int \psi\,\mathrm d\tau = \int x \psi x^{-1}\,\mathrm d\tau'$$ But the antisymmetry means that:

$$x\psi x^{-1} = -\psi$$

It shouldn't matter what basis we use for the integral. The transformation $x$ preserves inner products.

But I think this means that: $$\int \psi\,\mathrm d\tau = \int x \psi x^{-1}\,\mathrm d\tau' = \int (-1)\psi \,\mathrm d\tau' = -\int \psi\,\mathrm d\tau$$

Which implies the integral is zero. Now, if the representation contains the totally symmetric representation, then antisymmetric element $x$ does not exist, which invalidates the rationale above, though the other components of the integral are zero.

EDIT: Made this more rigorous. (I think. Feel free to critique).

First, let's decompose $\psi$ into components that correspond to how its representation reduces. That is, if

$$\Gamma_{\psi} = \Gamma_{1}\oplus\Gamma_{2}+\ldots$$

we write

$$\psi = \phi_{1} + \phi_{2} + \ldots $$

where the symmetry of $\phi_{i}$ is captured by the $\Gamma_{i}$.

Suppose $\Gamma_{i}$ is not fully symmetric. There is some symmetry element $x$ whereby $x$ is antisymmetric in this representation. In other words, $$x\phi_{i} = -\phi_{i}$$

Now, consider this operation. You could perform the change of variable $x\tau \rightarrow \tau'$ for the integral.

$$\int \phi_{i}\,\mathrm d\tau = \int x \phi_{i} x^{-1}\,\mathrm d\tau'$$

This step is straightforward because we're basically arguing that if we convert our basis to something else (spcifically, transformed under $x$), the integral will be the same. This must be true since rotating the coordinate system, for example, does not change the value of the integral.

Now, we can substitute our definitions above:

$$\int \phi_{i}\,\mathrm d\tau = \int x \phi_{i} x^{-1}\,\mathrm d\tau' = \int (-1)\phi_{i} \,\mathrm d\tau = -\int \phi_{i}\,\mathrm d\tau$$

where I have used $x^{-1}\tau' = \tau$.

The equality means that the integral must be zero. 

Now, if the irreducible representation is the totally symmetric representation, then antisymmetric element $x$ does not exist, which invalidates the rationale above, though the other components of $\psi$ will integrate to zero.

This argument works quite well when the irreducible representation $\Gamma_{i}$ is 1-dimensional. What happens when $\Gamma_{i}$ is higher dimensional?

I'm not sure I can rigorously prove this part, but I think I can sketch it out for a simple example.

Consider point group $C_{3}$.

\begin{array}{|c|c|c|} \hline & E & 2C_{3} \\ \hline A & 1 & 1 \\ E & 2 & -1 \\ \hline \end{array}

However, you could also write this character table as:

\begin{array}{|c|c|c|} \hline & E & C_{3} & C_{3}^{-1} \\ \hline A & 1 & 1 & 1 \\ E_{1} & 1 & \varepsilon & \varepsilon^{*} \\ E_{2} & 1 & \varepsilon^{*} & \varepsilon \\ \hline \end{array}

where $\varepsilon = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$. Note that a linear combination of $C_{3}$ and $C_{3}^{-1}$, namely $\frac{\sqrt{2}}{2}C_{3} + \frac{\sqrt{2}}{2}C_{3}^{-1}$, provides a linear combination that is equal to operation $-E$. Wait, you ask, but that is not an element of the $C_{3}$ point group. That's right, but a $p_{x}$ orbital does not have 3-fold symmetry either. But if you rotate it both $120^{\circ}$ and $-120^{\circ}$, you get two orbitals that when mixed, gives the $p_{x}$ orbital rotated $180^{\circ}$.

I think this generalizes out to other 2-dimensional and higher dimensional irreducible representations in that you either have a matrix that inverts (for example character -2 in a 2-dimensional representation) or you are guaranteed that a linear combination of representations gives the inversion (like here). My background here is too weak at present to provide anything more concrete.

This might not be as rigorous as desired. Hopefully, it's correct.

Suppose $\Gamma_{\psi}$ is not fully symmetric. There is some symmetry element $x$ whereby $x$ is antisymmetric in this representation.

Now, consider this operation. You could perform the change of variable $x\tau \rightarrow \tau'$ for the integral, and I believe this requires a coordinate change on $\psi$ as well.

$$\int \psi d\tau = \int x \psi x^{-1}d\tau'$$ But the antisymmetry means that:  $$x\psi x^{-1} = -\psi$$

It shouldn't matter what basis we use for the integral. The transformation $x$ preserves inner products.

But I think this means that: $$\int \psi d\tau = \int x \psi x^{-1}d\tau' = \int (-1)\psi d\tau' = -\int \psi d\tau$$

Which implies the integral is zero. Now, if the representation contains the totally symmetric representation, then antisymmetric element $x$ does not exist, which invalidates the rationale above, though the other components of the integral are zero.

This might not be as rigorous as desired. Hopefully, it's correct.

Suppose $\Gamma_{\psi}$ is not fully symmetric. There is some symmetry element $x$ whereby $x$ is antisymmetric in this representation.

Now, consider this operation. You could perform the change of variable $x\tau \rightarrow \tau'$ for the integral, and I believe this requires a coordinate change on $\psi$ as well.

$$\int \psi\,\mathrm d\tau = \int x \psi x^{-1}\,\mathrm d\tau'$$ But the antisymmetry means that: 

$$x\psi x^{-1} = -\psi$$

It shouldn't matter what basis we use for the integral. The transformation $x$ preserves inner products.

But I think this means that: $$\int \psi\,\mathrm d\tau = \int x \psi x^{-1}\,\mathrm d\tau' = \int (-1)\psi \,\mathrm d\tau' = -\int \psi\,\mathrm d\tau$$

Which implies the integral is zero. Now, if the representation contains the totally symmetric representation, then antisymmetric element $x$ does not exist, which invalidates the rationale above, though the other components of the integral are zero.