1. General oxidation state of lanthanoids

   The most common oxidation state of lanthanoids is $+3$ because they usually lose two $6s$ electrons and one $5d/4f$ electron.

2. When do lanthanoids deviate from $+3$?

   Deviation from $+3$ occurs when some other oxidation state gives an especially stable electronic configuration, such as:

   • empty $4f$ subshell: $4f^0$
   • half-filled $4f$ subshell: $4f^7$
   • fully filled $4f$ subshell: $4f^{14}$

3. Check the given elements

   A: Ce $(Z=58)$

   Cerium has configuration approximately: $[Xe]4f^1 5d^1 6s^2$ In the $+3$ state: $Ce^{3+} : [Xe]4f^1$ In the $+4$ state: $Ce^{4+} : [Xe]4f^0$ Since $4f^0$ (empty $f$ subshell) is especially stable, cerium readily shows $+4$ in addition to $+3$.

   So, Ce easily deviates from $+3$.

   B: La $(Z=57)$

   $La : [Xe]5d^1 6s^2$ $La^{3+} : [Xe]$ Lanthanum strongly prefers $+3$ and does not commonly show other oxidation states.

   C: Lu $(Z=71)$

   $Lu : [Xe]4f^{14}5d^1 6s^2$ $Lu^{3+} : [Xe]4f^{14}$ The $+3$ state itself gives the very stable fully filled $4f^{14}$ configuration, so it does not tend to deviate easily.

   D: Gd $(Z=64)$

   $Gd : [Xe]4f^7 5d^1 6s^2$ $Gd^{3+} : [Xe]4f^7$ The $+3$ state gives the very stable half-filled $4f^7$ configuration, so it also prefers $+3$.

4. Conclusion

   Among the given options, cerium most readily deviates from the common $+3$ oxidation state because $Ce^{4+}$ attains the stable $4f^0$ configuration.

$\boxed{\text{A: Ce}}$