Yes. The potential $V$ is a linear function of $Q$ (due to the Coulomb law), so the charge $Q$ can be factored out on both sides of the equation.
To find the capacitance for given metal body, we express it as
$$
C= \frac{Q}{V_0},
$$
where $V_0$ is the potential on the surface. To get $C$, we need to find $Q$ on the surface. This amount to solving Laplace equation with given boundary conditions - potential $V_0$ on the surface and 0 at infinity.
Due to the Coulomb law, we can write the potential anywhere as surface integral:
$$
V(\mathbf x) = \frac{1}{2}K \int_S \frac{\sigma(\mathbf x')}{|\mathbf x - \mathbf x '|} d^2 \mathbf x'.
$$
For $N$ points on the surface $\mathbf x_i$, this can be discretized into system of linear equations for $\sigma_i = \sigma (\mathbf x_i)$ :
$$
V_0 = \sum_{j=1}^N K_{ij} \sigma_j,~~~~i=1,2,...,N
$$
where
$$
K_{ij} = \frac{\frac{1}{2}K}{|\mathbf x_i- \mathbf x'_j|}\Delta S_{j}
$$
Explicit expression for $\sigma_j$ would look horribly, because we need to use large $N$ even for simplest surfaces. Solve for $\sigma_i$ numerically; the total charge is then
$$
Q = \sum_{j=1}^N \sigma_j \Delta S_j.
$$
Substitute in the first formula and you have the capacitance.
