More than one photon can be absorbed, but the probability is minute for usual intensities. As a scale for "usual intensities" note that sunlight on earth has an intensity of about $1000\,\mathrm{W/m^2} = 10^{-1}\mathrm{W/cm^2}$.
The intuitive reason is, that the linear process (an electron absorbs one photon) is more or less "unlikely" (as the coupling between the em. field and electrons is rather weak), so a process where two photons interact is "unlikely"$^2$ and thus strongly suppressed. So for small intensities the linear process will dominate distinctly. The question is only, at what intensities the second order effects will become visible.
In the paper by Richard L. Smith, "Two-Photon Photoelectric Effect", Phys. Rev. 128, 2225 (1962) the photocurrent for radiation above half of the cutoff frequency but below the cutoff frequency is discussed. They note that for usual intensities the photocurrent will be minute, but that given strong enough fields
such as those observed in a focus spot of a laser (on the order of $10^7\,\mathrm{W/cm^2}$) the effect might be measurable. They also note, that thermal heating by the laser field may make the pure second order effect unobservable.
The more recent paper S. Varró, E. Elotzky, "The multiphoton photo-effect and
harmonic generation at
metal surfaces", J. Phys. D: Appl. Phys. 30, 3071 (1997) dicusses the case where high intensities (on the scale of $10^{10}\,\mathrm{W/cm^2}$) produce even higher order effects (and unexpectedly high, coherent non-linear effects, that is absorption of more than two photons by one electron). Their calculations explain the experimental observations of sharp features in the emission spectra of metal surfaces.
Historical fun fact: The 1962 paper is so old, that it talks about an "optical ruby maser"; lasers where so new back then, they did not even have their name yet.
There are a couple of reasons for this. First and foremost, the electrons are ejected from the surface of the metal in random directions. When you measure things like the "stopping potential" you're only sensitive to motion in directions that would carry the photo-electron to the anode. Because you're only sensitive to motion in particular directions, you only see the part of the kinetic energy along that particular direction. It is this kinematic messiness that makes measuring the reverse potential needed to stop all current the preferred method of measuring the photoelectric effect - the last electrons stopped will be the ones where the largest possible fraction of the photon's energy went in to propelling the electron toward the anode.
Secondarily, the valence electrons in metals have energies in what are known as the "conduction band", which means the electron's energy can exist in a continuum. That fact, combined with the random messiness inherent in thermodynamics, means that no two electrons will have the same kinetic energy before the photons hit them. Thus, each electron will have a slightly different kinetic energy after it gets ejected from the metal.
Of approximately equal weight is the spread in frequencies for the incident light. See, even if your light is produced by a nice sharp atomic line, like in a low pressure mercury lamp, the mercury atoms in the gas will be undergoing thermal motion, leading to Doppler broadening of the line.
Best Answer
The answer to this particular question is simply, unequivocally, no.
It is inconceivable that all the electrons could be removed while leaving something recognizable as the metal (object). Recall that it is the (outermost, valence) electrons that determine chemical properties.
If hypothetically, by some unknown mechanism, all of the electrons were removed, only the positively charged nuclei of the atoms would remain, all (strongly) repelling each other and with no way to form chemical bonds.
Other answers have addressed the fact that the metal becomes positively charged as electrons are ejected and collected thus attracting electrons back to the metal so this answer is just to address the quoted question.