A disadvantage of using cubical singular homology comes form the following example (copied from my daughter's e-mail):

Example. The circle $c(t)=(\cos 2\pi t,\sin 2\pi t)$, $t\in [0,1]$ is not the boundary of any singular cubical chain.

Proof. For a singular chain $\sum_i a_i c_i$, where $a_i\in\mathbb{Z}$ and $c_i$ is a singular cube, define its `length' by $\sum_i a_i$. Since opposite sides of a singular cube have signs $\pm 1$, it follows that the length of $\partial c_i$ equals zero and hence the length of $\partial(\sum_i a_i c_i)$ equals zero. However, the length of $c(t)=(\cos 2\pi t,\sin 2\pi t)$ equals $1$ so it cannot be represented as a boundary of a singular chain. $\Box$

If you consider the singular cube: $b=(s\cos 2\pi t, s \sin 2\pi t)$, $s,t\in [0,1]$, then $\partial b$ is a difference of two singular cubes: one is $c$ and one is a constant cube (mapping to the center). It is not possible to avoid such degenerate cubes.

As I understand this problem is what Tyler Lawson had in mind when he wrote in his answer: The main disadvantage that comes solely from the point of view of homology is that you have this irritating normalization procedure: you have to take the cubical chains on X and take the quotient by degenerate cubes.

A disadvantage of using cubical singular homology comes form the following example (copied from Natalia Hajlasz's homework):

Example. The circle $c(t)=(\cos 2\pi t,\sin 2\pi t)$, $t\in [0,1]$ is not the boundary of any singular cubical chain.

Proof. For a singular chain $\sum_i a_i c_i$, where $a_i\in\mathbb{Z}$ and $c_i$ is a singular cube, define its `length' by $\sum_i a_i$. Since opposite sides of a singular cube have signs $\pm 1$, it follows that the length of $\partial c_i$ equals zero and hence the length of $\partial(\sum_i a_i c_i)$ equals zero. However, the length of $c(t)=(\cos 2\pi t,\sin 2\pi t)$ equals $1$ so it cannot be represented as a boundary of a singular chain. $\Box$

If you consider the singular cube: $b=(s\cos 2\pi t, s \sin 2\pi t)$, $s,t\in [0,1]$, then $\partial b$ is a difference of two singular cubes: one is $c$ and one is a constant cube (mapping to the center). It is not possible to avoid such degenerate cubes.

As I understand this problem is what Tyler Lawson had in mind when he wrote in his answer: The main disadvantage that comes solely from the point of view of homology is that you have this irritating normalization procedure: you have to take the cubical chains on X and take the quotient by degenerate cubes.