What is the strongest oxidising agent?
Ivan's answer is indeed thought-provoking. But let's have some fun.

IUPAC defines oxidation as: 


  The complete, net removal of one or more electrons from a molecular
  entity.


My humble query is thus - what better way is there to remove an electron than combining it with a literal anti-electron? Yes, my friends, we shall seek to transcend the problem entirely and swat the fly with a thermonuclear bomb. I submit as the most powerful entry, the positron.

Since 1932, we've known that ordinary matter has a mirror image, which we now call antimatter. The antimatter counterpart of the electron ($\ce{e-}$) is the positron ($\ce{e+}$). To the best of our knowledge, they behave exactly alike, except for their opposite electric charges. I stress that the positron has nothing to do with the proton ($\ce{p+}$), another class of particle entirely.

As you may know, when matter and antimatter meet, they release tremendous amounts of energy, thanks to $E=mc^2$. For an electron and positron with no initial energy other than their individual rest masses of $\pu{511 keV c^-2}$ each, the most common annihilation outcome is:

$$ \ce{e- +\ e+ -&gt; 2\gamma}$$

However, this process is fully reversible in quantum electrodynamics; it is time-symmetric. The opposite reaction is pair production:

$$ \ce{2\gamma -&gt; e- +\ e+ }$$

A reversible reaction? Then there is nothing stopping us from imagining the following chemical equilibrium:

\begin{align}
\ce{e- +\ e+ &amp;&lt;=&gt; 2\gamma} &amp;
\Delta_r G^\circ &amp;= \pu{-1.022 MeV} =\pu{-98 607 810 kJ mol^-1}
\end{align}

The distinction between enthalpy and Gibbs free energy in such subatomic reactions is completely negligible, as the entropic factor is laughably small in comparison, in any reasonable conditions. I am just going to brashly consider the above value as the standard Gibbs free energy change of reaction. This enormous $\Delta_r G^\circ$ corresponds to an equilibrium constant $K_\mathrm{eq} = 3 \times 10^{17276234}$, representing a somewhat product-favoured reaction. Plugging the Nernst equation, the standard electrode potential for the "reduction of a positron" is then $\pu{+2 116 413 V}$. 

Ivan mentions in his answer using an alpha particle as an oxidiser. Let's take that further. According to NIST, a rough estimate for the electron affinity of a completely bare darmstadtium nucleus ($\ce{Ds^{110+}}$) is $\pu{-204.4 keV}$, so even a stripped superheavy atom can't match the oxidising power of a positron!

... that is, until you get to $\ce{Ust^{173+}}$ ...