It's a nice question. There is indeed a way to formulate an unconditional analogue. The idea is to no longer interpret $\gamma$ as ordinates, but instead label the zeroes as $\tfrac{1}{2}+i\gamma$, where $\gamma$ is a complex number with imaginary part in between -1/2 and 1/2. In this way, the Riemann hypothesis is the statement that all $\gamma$ are real, and, on RH, these $\gamma$'s correspond exactly to your $\gamma$'s. One can in this case show unconditionally that

$$
\lim_{t\rightarrow\infty} \frac{1}{N(T)}\sum_{T<\Re(\gamma),\Re(\gamma')<2T} T^{i\alpha(\gamma-\gamma')}\frac{4}{4+(\gamma-\gamma')^2} = \alpha
$$

for fixed $0 < \alpha < 1$. (If you want this uniformly for $\alpha$ you'll have to restrict $\alpha$ away from $1$ and $0$, or modify things around $0$.) Note that $\gamma-\gamma'$ may be complex.

Observations of this sort go back at least to Rudnick and Sarnak's paper "Zeros of principal L-functions and random matrix theory," where they write of the k-point correlation sums

$$
\sum_{\gamma_1,\gamma_2,...\gamma_k \textrm{ distinct}}h\Big(\frac{\gamma_1}{T}\Big)\cdots h\Big(\frac{\gamma_k}{T}\Big)f(\frac{L}{2\pi}\gamma_1,...,\frac{L}{2\pi}\gamma_k\Big)
$$

"if $h$ and $f$ are defined for complex argument and are localized, then the sums make sense even if we do not assume RH." This means that $f(x+iy) = \int e((x+iy)\xi)\hat{f}(\xi)d\xi$.

The reason one can do this here and with the slightly different 2-point sum Montgomery considers is that the explicit formulas relating zeroes to primes that Montgomery uses in his proof remain true unconditionally so long as the $\gamma$ representing an ordinate of a zero is replaced by $\gamma$ as defined above. I don't know that, for the Montgomery 2-point sum, this argument appears explicitly in print anywhere, but it's an instructive exercise to go through the standard proof seeing that everything still works and you can arrive at the formulation I have above.

Recalling the symmetry of the zeros (that is, if $\tfrac{1}{2}+i\gamma$ is a zero, then so are $\tfrac{1}{2}-i\gamma$, $\tfrac{1}{2}+i\bar{\gamma}$, and $\tfrac{1}{2}-i\bar{\gamma}$), one can see that $F(\alpha)$ (if this limit exists) is positive and symmetric using the same argument that Montgomery does. (Positivity requires a slight modification -- it is here that one needs conjugate symmetry in $\gamma$; I can elaborate if you'd like.)

I don't think anything can be said about the zeros that are already known to lie on the critical line, because in particular I don't think anything explicit can be said about the relation between these zeroes and the primes (about which we can say at least a few useful things). Using a zero-density estimate, it may be possible to show that

$$
\lim_{t\rightarrow\infty} \frac{1}{N(T)}\sum_{T<\Re(\gamma),\Re(\gamma')<2T} T^{i\alpha(\Re(\gamma-\gamma'))}\frac{4}{4+(\Re(\gamma-\gamma'))^2} = \alpha
$$

for $\alpha$ restricted as above, even if RH is false, returning our attention once again to the ordinates of zeroes. This would be interesting (to me) for a variety of reasons, but without a clever idea I don't think any known zero-density estimates are enough to get here.