Nope. From the Hubble FAQ:

Given the latest advances in Hubble instrumentation, image processing etc., what would the absolute theoretical limit for the smallest object visible on the Moon be?

Firstly we should say that a bright, high-contrast feature such as a star can be seen however small (in angular terms) it appears. In these cases the star would just appear as a dot. So, if there were a very shiny surface on the Moon that caught the Sun, it might be seen from Earth with quite a small telescope.

Here we will try to answer the related question of how close together two features can be and still be discerned as separate – this is called the angular resolution. The Rayleigh criterion gives the maximum (diffraction-limited) resolution, R, and is approximated for a telescope as

R = λ/D, where R is the angular resolution in radians and λ is the wavelength in metres. The telescope diameter, D, is also in metres.

In more convenient units we can write this as:

R (in arcseconds) = 0.21 λ/D, where λ is now the wavelength in micrometres and D is the size of the telescope in metres.

So for Hubble this is:

R = 0.21 ｘ 0.500/2.4 = 0.043 arcseconds (for optical wavelengths, 500 nm) or

R = 0.21 ｘ0.300/2.4 = 0.026 arcseconds (for ultraviolet light, 300 nm).

Note that the resolution gets better at shorter wavelengths, so we will use the second of these numbers from now on.

Hubble’s optics are now essentially perfect, and the telescope is above the Earth’s atmosphere, so this gives an accurate value for the resolution of the image produced by the telescope before it is captured by one of the telescope’s detectors. However, the detectors have pixels that are quite large relative to these values in most cases and this degrades the resolution somewhat. The pixels of Hubble’s latest UV-sensitive instrument, the UVIS channel of the Wide Field Camera 3 are 0.04 arcseconds across. This means that the final effective resolution of telescope and detector can be estimated as:

R = √(telescope optical resolution2 + pixel size2)

so for Hubble with WFC3/UVIS, and in the UV (300 nm) we get

R = √(0.0262 + 0.0402) = 0.048 arcseconds

Then, in an extreme case, such as the Moon, where there is lots of light (a high signal/noise ratio), it’s possible to do image processing (image restoration) and retrieve roughly a factor of two better resolution at the expense of some artefacts. So for Hubble, we conclude that the best resolution we are likely to manage is about 0.024 arcseconds (in the ultraviolet). On the Moon, at its closest distance to the Earth, this would give a linear resolution of:

363 000 000 ｘ R /206 000 = 43 metres

So the minimum separation of two objects on the Moon that could be seen as separate when observed by Hubble in the ultraviolet should be about 40 metres. Unfortunately it is very difficult for Hubble to observe the Moon — because the telescope is rapidly orbiting the Earth the Moon appears to swing backwards and forwards in the sky very rapidly and it is almost impossible for the telescope to compensate — so it is unlikely that this limit could ever be approached.