Although increasing an object's spin will increase its total energy, it will not increase its tendency to collapse to a black hole. The physical reason for this, is that conservation of angular momentum will work against any possible collapse. From a slightly different perspective, it can also be seen from the fact that the event horizon radius of a spinning black hole is smaller than that of a non-spinning black hole of the same mass.
A practical and astrophysically relevant consequence of this is that rotating neutron stars can be more massive than non-rotating neutron stars. This allows for the reverse to the OPs scenario to happen:
Suppose a neutron star is created with a very high spin (maybe from the collision of two other neutron stars), and a mass that is higher than the critical mass for a non-rotating neutron star to collapse to a black hole. Initial the neutron star is kept stable by its angular momentum, but over time it will lose angular momentum (e.g. due to emission of EM radiation) and spin down. 43
At some point the angular momentum will be insufficient to prevent collapse, and the neutron star collapses to a black hole.
This leaves the question what portion of a rotating black hole's mass can be thought off as consisting of "rotational energy". This is not straightforward to answer since in general relativity there is no clear cut separation of different kinds of energy. However, some indication can be gleamed from looking at the rotational energy of neutron stars at the critical point of collapse. Table II of arXiv:1905.03656 gives values for the mass ($M$), angular momentum ($J$), and rotational energy ($T$) of such neutron stars depending on the model for the equation of state for the neutron star. For one such model these values are
\begin{align} M & = 2.57 M_{\odot} \ J &= 4.183\times10^{49} \text{erg s}\ T &= 2.415 \times 10^{53} \text{erg} \end{align}
This translates to a spin parameter
$$\chi = \frac{c J}{GM^2} = 0.719,$$
i.e. it would collapse to a black hole spinning at 72% of its maximum rate. However, the fraction of its total energy in rotational energy ($T/(Mc^2)$) is only about 5 percent. 150