Is spectral leakage due to windowing 'different' for the DTFT and DFT?
The concept of spectral leakage is not dependent on the context, and it's the same thing for the DTFT as it is for the DFT. It's probably helpful to remember that the DFT of a finite length sequence equals a sampled version of the DTFT of the same sequence:

$$\begin{align}X(\omega)&amp;=\sum_{n=0}^{N-1}x[n]e^{-jn\omega}\qquad\text{(DTFT)}\\
\tilde{X}[k]&amp;=\sum_{n=0}^{N-1}x[n]e^{-jn\frac{2\pi k}{N}}\qquad\text{(DFT)}\end{align}$$

Obviously we have

$$\tilde{X}[k]=X\left(\frac{2\pi k}{N}\right)$$

Now assume we have two sinusoidal signals windowed with a rectangular window such that in one case there's an integer number of periods inside the window, and in the other cases there isn't:



In both cases we get spectral leakage due to the windowing process. However, in the first case, the samples of the DTFT computed by the DFT are all at the zeros of the DTFT (except for one sample at the sinusoid's frequency). This is not the case for the second signal. That means that in the first case the DFT doesn't "see" the spectral leakage due to the sampling process, but in the other case the spectral leakage is also visible in the DFT.

These two cases are shown in the figure below. The DTFT is shown in blue, and the corresponding DFT is in red: