= Solution
Use the <Green function of killed planar Brownian motion>, normalized as the density of expected occupation with respect to area:
$$
\mathbb E_z\int_0^T f(B_t)\,dt
=\int_D G_D(z,w)f(w)\,dA(w)
$$
for nonnegative measurable $f$. Equivalently, $G_D(z,w)=\int_0^\infty p_D(t,z,w)\,dt$, where $p_D$ is the <killed Brownian transition density>. With generator $\tfrac12\Delta$, the distributional normalization is $-\tfrac12\Delta_wG_D(z,w)=\delta_z(w)$, and the singularity is $-\pi^{-1}\log|w-z|$ plus a locally <harmonic function>. This fixes the normalization of the <Dirichlet Green function> explicitly.
By <conformal invariance of planar Brownian motion> and its <conformal Brownian clock>,
$$
\mathbb E_{z'}\int_0^{T'} f(B'_s)\,ds
=\mathbb E_z\int_0^T f(\phi(B_t))|\phi'(B_t)|^2\,dt
=\int_D G_D(z,w)f(\phi(w))|\phi'(w)|^2\,dA(w).
$$
The <Jacobian determinant> of a <conformal map> is $|\phi'(w)|^2$. Changing the area variable to $w'=\phi(w)$ gives
$$
\int_{D'}G_D(z,\phi^{-1}(w'))f(w')\,dA(w').
$$
Uniqueness of the occupation density proves the desired equality almost everywhere. Both functions are continuous and <harmonic> away from their pole, so it holds at every $w\ne z$. \b[The Green function is conformally invariant]:
$$
\boxed{G_{D'}(\phi(z),\phi(w))=G_D(z,w).}
$$
If the <Dirichlet Green function> is instead normalized for $-\Delta$, both kernels are divided by two and the invariance statement is unchanged.
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