Everything I'm going to say below (and much more) is well-known but I still decided to demonstrate one crude elementary technique that often allows to get the correct range of $p,q$ for such kind of operators except for endpoints. I'll leave the endpoint discussion to someone cleverer and more knowledgeable than I.
First of all, the condition (2) should really be $1/p-1/q=(1-\alpha)$. That is a smoothing operator, i.e., it makes functions less singular at a point but more singular at $\infty$, so definitely $q$ should be greater than $p$, not the other way around. Also you should add that $1<p,q<+\infty$ in that boundedness statement.
Now let's discern some simple necessary conditions coming ultimately from the fact that the phase doesn't really help with anything when $|xy|\le 1$.
Take $\delta<1$ and start with $f=\delta^{-1/p}\chi_{[0,\delta]}$ so $\|f\|_p=1$ and look only at $x\in[0,\delta]$. Then $|Tf(x)|\approx \delta^{-1/p}\delta^{1-\alpha}$, so
$$
\int_0^\delta |(Tf)(x)|^q\approx \delta\left[\delta^{-1/p}\delta^{1-\alpha}\right]^q=
\delta^{q[\frac 1q-1/p+1-\alpha]}
$$
so not to blow it up for $\delta\to 0$, we must have $1/q-1/p+(1-\alpha)\ge 0$, i.e., $1/p-1/q\le 1-\alpha$ (which is not surprising because it is the same scaling only in one direction now).
Next, notice also that we have $(Tf)(x)\approx \delta^{-1/p}\delta x^{-\alpha}\ge \delta^{-1/p}\delta^{1+\alpha}$ for $\frac1{2\delta}\le x\le \frac 1\delta$. Thus we get
$$
\int_{\frac 1{2\delta}}^{\frac 1\delta}|Tf|^q\ge \tfrac 12\delta^{-1} [\delta^{-1/p}\delta^{1+\alpha}]^q=\delta^{q[-\frac 1q-\frac 1p+1+\alpha]}
$$
so not to blow this one up for $\delta\to 0$, we must have $\frac 1p+\frac 1q\le 1+\alpha$.
Now we can play with large $\delta$ and the values of $Tf$ on $[0,1/\delta]$.
We get $|Tf(x)|\approx \delta^{-1/p}\delta^{1-\alpha}$, so
$$
\int_0^{1/\delta}|Tf|^q\approx \delta^{-1}[\delta^{-1/p}\delta^{1-\alpha}]^q
=\delta^{q[-\frac 1q-\frac 1p+1-\alpha]}
$$
so, not to blow that one up as $\delta\to\infty$, we should have $\frac 1p+\frac 1q\ge 1-\alpha$.
Last, as Piero observed, this operator is equivalent to the convolution operator with the kernel $\frac{e^{ix^2/2}}{|x|^\alpha}$. Convolution operators with non-zero decaying kernels can never act from $L^p$ to $L^q$ unless $q\ge p$. Indeed, if $K$ is the kernel and $f$ is some bounded bump on $[0,1]$ for which $g=K*f$ is not zero, then we can consider $F(x)=\sum_{j=1}^N f(x-u_j)$, so $G(x)=(TF)(x)=\sum_{j=1}^Ng(x-u_j)$. When $u_j$ are very far apart, the summands have morally disjoint supports in both cases, so we have $\|F\|_p=N^{1/p}\|f\|_p$ and $\|G\|_q=N^{1/q}\|g\|_q$, i.e., not to blow the norm up for large $N$, we must have $1/q\le 1/p$, i.e., $q\ge p$.
Thus the conditions
$$
0\le \frac 1p-\frac 1q\le 1-\alpha,\qquad 1-\alpha\le \frac 1p+\frac 1q\le 1+\alpha
$$
are necessary. Let's show now that if the second set of inequalities here is strict, then the operator is bounded from $L^p$ to $L^q$.
To this end, use the restricted Whitney decomposition $\mathcal Q$ of the plane using dyadic squares $Q=I\times J$ of size $1$ near and on the diagonal $D=\{x=y\}$ and dyadic squares of size about $0.1{\rm dist}(Q,D)$ away from the diagonal. Write $Tf=\sum_a T_af$ where $a=1,2,4,8,\dots$ and
$$
T_af=\sum_{Q=I\times J\in\mathcal Q:|I|=|J|=a} \chi_I T(f\chi_J)\,.
$$
Notice that each $T_a$ is essentially a block operator, i.e., for each $I$ only fixed number of $J$ is used and each $J$ is used only with fixed number of $I$'s. The norm of a block operator from $L^p$ to $L^q$ when $q\ge p$ is essentially the maximum of the norms of single cells.
For cells of size $1$ we just use the trivial absolute value bound on the kernel to get some constant norm from the condition $\frac 1p-\frac 1q\le(1-\alpha)$. Now assume that the cell $Q=I\times J$ has size $a>1$, so the distance from that cell to the diagonal is at least $10a$.
If $x_I, y_J$ are the centers of $I,J$ and $x=x_I+x'\in I, y=y_J+y'\in J$, then you can write (denoting $A=|x_I-y_I|\ge 10a$
$$
|x-y|^{-\alpha}=A^{-\alpha}\sum_{k,\ell}c_{k,l}(x'/A)^k (y'/A)^\ell
$$
with $\sum_{k,\ell}|c_{k,l}|\|x'/A\|_{\infty}^k \|y'/A\|_{\infty}^\ell\le C<+\infty$
independently of $a$. I shamelessly use real analyticity here though what is really needed for such representation is just the estimate for partial derivatives of order 2 or 3. To cover the latter case, one merely needs to use the Fourier decomposition instead of Taylor.
Thus, it is enough to estimate the norm of the classical Fourier transform times $a^{-\alpha}$ acting from $L^p(J)$ (functions supported on $J$) to $L^q(I)$. If we get $a$ to some negative power, we will be able to sum up $T_af$ in $L^q$ using Minkowski's inequality and finish the story.
The latter is classical indeed. Just use the $L^2$ bound and $L^p\to L^{p'}$ bound for $p\le 2$. For completeness here is the casework. Let $\|f\|_{L^p(J)}=1$
$p\ge 2$. Then $q\ge 2$ as well. $\|f\|_2^2\le a^{1-\frac2p}$ and $\|f\|_{L^1(J)}\le a^{1-\frac 1p}$ by Holder. Hence
$$
a^{-\alpha q}\|\widehat f\|_{L^q(I)}^q\le a^{-\alpha q}a^{1-\frac 2p}a^{(q-2)(1-\frac 1p)}=a^{q[-\alpha+\frac 1q-\frac{2p}q+1-\frac 1p-\frac 2q+\frac {2p}q]}
=a^{q[-\frac 1p-\frac 1q+(1-\alpha)}
$$
and the condition $\frac 1p+\frac 1q\ge (1-\alpha)$ finishes the story.
$p<2, q>p'$. Then we use $L^p\to L^{p'}$ action instead of $L^2$ to $L^2$ and write
$$
a^{-\alpha q}\|\widehat f\|_{L^q(I)}^q\le
a^{-\alpha q}\|\widehat f\|_{L^{p'}(I)}^{p'}\|\widehat f\|_\infty^{q-p'}\le
a^{-\alpha q}\cdot 1\cdot a^{(q-p')(1-1/p)}=a^{q[-\alpha+1-\frac 1p-\frac 1q]}
$$
and the condition $\frac 1p+\frac 1q>(1-\alpha)$ finishes the story.
$p<2, q\le p'$. Then we still use $L^p\to L^{p'}$ but Holder on the Fourier side:
$$
a^{-\alpha q}\|\widehat f\|_{L^q(I)}^q\le
a^{-\alpha q}a a^{-\frac q{p'}}=
a^{q[-\alpha+\frac 1q-1+\frac 1p]}
$$
and the condition $\frac 1p+\frac 1q<1+\alpha$ can be used.
As I said, I leave it to somebody else (Terry Tao?) to explain what happens at the endpoint cases but for now just use this outline to show that the same range (with strict inequalities) is fine for any kernel $e^{ixy}K(x,y)$ where $K(x,y)$ satisfies $|\partial^{\beta}K|\le |x-y|^{-\alpha-|\beta|}$ for $0\le |\beta|\le 4$, say. That should be a good exercise to check if you understand everything here. If not, feel free to ask as many questions as you want :-)
