No; the equator is always warmer than the poles
Earth has a 23 degree axial tilt. That means that the sun is never more than 23 degrees from the equator. The equation for incident solar energy ($E$) for any given day is
$$ E = E_d\cos{A_i}$$ where $E_d$ is the incident energy of the sun overhead at noon and $A_i$ is the angle of the the sun above the horizon at noon. There are plenty of associated factors like cloud cover and refraction of light from various layers of the atmosphere, but we can ignore them for now.
On the Earth, over one year of sunlight, the total incident sunlight can be calculated as the percentage of the maximum possible sunlight ($E_{max}$); that is, the amount of energy you would receive if the sun was directly overheat at noon every day of the year (i.e. a planet with no axial tilt). We will define a year to be $2\pi$ units long, so $E_{max} = 2\pi E_d$.
The motion of the sun over the course of the year can be modeled by the sine (or cosine) function as $$A_i = \max\left[\frac{\pi}{2}, A_{tilt}\sin{t} + L\right]$$ where $t \in [0, 2\pi]$ are times within one solar year and $A_{tilt}$ is the axial tilt of the planet and and $L$ is the absolute value of latitude (north and south do not matter). Note that if $L$ is greater than $A_{tilt}$, then the location is outside of the tropics and the sun can never be overhead. Also note the max function is necessary, since if the sun is more than 90 degrees away, it is still giving zero light.
Plugging equation 2 into equation 1 and integrating over a year, we get
$$
\begin{align}
\frac{E}{E_{max}} &= \frac{1}{2\pi}\int_{0}^{2\pi}\cos\left(\max\left[\frac{\pi}{2}, A_{tilt}\sin{t} + L\right]\right)dt\\
\end{align}
$$
The closed form solution of this indefinite integral is derived from the Bessel function with some added complexities due to the max function, but we can solve it numerically. For Earth, axial tilt is 23.5 degrees or 0.410 radians, and the solution is roughly 0.958. That is, the equator gets 95.8% as much sunlight as a planet with no axial tilt. A solution for the Tropic of Cancer (or Capricorn) at 23.5 degrees from the equator is 0.878. At 60 degrees (north or south) the value is 0.478, while at the pole it is 0.128. So far, this reflects reality pretty well.
Now, let us change the axial tilt of your planet to 60 degrees. The corresponding number for the equator is 0.743; and for the poles 0.294. You have succeeded in making the entire planet into a temperate or cooler climate; but you have not succeeded in making the poles warmer than the equator.
A general proof of this for any axial tilt can be seen by observing that the derivative of the integrated function is just the function itself. That function in turn is maximized by $L = 0$ for $A_{tilt} \in \left[0,\frac{\pi}{2}\right]$, so for any non-tidally locked planet orbiting the sun, the equator will receive the most solar radiation.
