José Figueroa-O'Farrill already answered the first question.
Let me answer the two other ones:

2) If $G$ is not $S^1$ or $S^3$, the answer is yes.
If this did not hold for some $\rho$, there would be a nonzero vector $a\in\mathbb C^{\dim\rho}$ that is in the kernel of $I(\rho,\gamma)$ for all $\gamma$.
Let $A$ be the square matrix whose every row is $a^*$.
Then $AI(\rho,\gamma)=0$ for all $\gamma$.

Now let $f:G\to\mathbb C$ be the unique smooth function that satisfies $\int_Gf(x)\rho(x)dx=A$ and $\int_Gf(x)\sigma(x)dx=0$ for all irreducible representations $\sigma$ that are not equivalent to $\rho$.
The existence and uniqueness of $f$ follows from the Peter-Weyl theorem.

For any $x\in G$ and any nontrivial homomorphism $\gamma:S^1\to G$ let
$$
F(x,\gamma)=\int_{S^1}f(x\gamma(t))dt
$$
be the integral of $f$ over the geodesic $t\mapsto x\gamma(t)$.
For any (unitary, irreducible) representation $\sigma$, a calculation gives
$$
\int_G F(x,\gamma)\sigma(x)dx
=
\int_G f(x)\sigma(x)dx I(\sigma,\gamma).
$$
By the definition of $f$, this is always zero (for both $\sigma=\rho$ and $\sigma\neq\rho$).
Thus $F$ is orthogonal to all matrix elements of all irreducible representations, so $F=0$.

This means that $f$ is not identically zero but its integral over every geodesic is zero.
This is impossible by theorem 1.1 of this paper.

If $G=S^1$, the answer is no for all nonconstant $\rho$.
If $G=S^3$, I don't know the exact answer.
On this group a function is known to integrate to zero over all geodesics if and only if it is antipodally antisymmetric ($f(-x)=-f(x)$ if we embed $S^3\subset\mathbb R^4$).
This with the previous argument might tell for which $\rho$s the answer is yes; it is not yes for all of them.

3) The answer to the third question is positive.

Let $D(\rho,\gamma)=\left.\frac{d}{dt}\rho(\gamma(t))\right|_{t=0}$.
Then $\frac{d}{dt}\rho(\gamma(t))=\rho(\gamma(t))D(\rho,\gamma)$.
Let $1$ denote the identity matrix, since $I$ is reserved for the integral.
If $A$ satisfies $AI(\rho,\gamma)=0$, then
\begin{eqnarray}
A
&=&
A(1-I(\rho,\gamma))
\\&=&
A\left(1-\int_{S^1}\rho(\gamma(t))dt\right)
\\&=&
A\int_{S^1}(\rho(\gamma(0))-\rho(\gamma(t)))dt
\\&=&
A\int_{S^1}\int_t^0\frac{d}{ds}\rho(\gamma(s))dsdt
\\&=&
A\int_{S^1}\int_t^0\rho(\gamma(s))dsdtD(\rho,\gamma).
\end{eqnarray}
If we let
$$
B=A\int_{S^1}\int_t^0\rho(\gamma(s))dsdt,
$$
then we have $A=BD(\rho,\gamma)$.
Moreover, the matrix $B$ can be explicitly expressed in terms of $A$.

Since $I(\rho,\gamma)$ is hermitean and $D(\rho,\gamma)$ skew-hermitean, the result states that the image of the derivative of a representation is exactly the kernel of the integral of the representation and vice versa.
(Here, of course, the derivative and the integral mean $D$ and $I$.)