There is no trouble at the end of the day, thanks to the fact that there are two, not just one, branch points at $p=0, \tfrac{1}{\alpha+1}$. One can take $[0,\tfrac{1}{\alpha+1}]$ to be the branch cut of the integrand in Eq. (3) and thereby make it analytic except along this cut. At the same time, the contour $\gamma$ should be chosen such that it encloses the whole branch cut.
Notice that in Eq. (2), the integral in $t$ is undefined for ${\rm Re}(\tfrac{1}{p})\ge\alpha+1$, and the branch cut falls within this region. It is consistent with our previous observation that the contour $\gamma$ should avoid the cut.
Now $\gamma$ can be deformed into an arbitrarily large circle. It certainly cannot smoothly pass through the simple pole at $p=1$ but has to leave behind a small clockwise contour around it. The contribution of this contour can be evaluated using the residue at the pole, and the result is $2\pi i \alpha^{-\nu-1}\Gamma(\nu+1)$. (I actually suspect that your definition of the Marcum Q-function is missing a factor of $1/2\pi i$.)
One still has to evaluate the integral over the large circle, which doesn't vanish. This can be done by making a change of variable $z = 1/p$. Then the integral will be turned into one over a contour encircling an essential singularity at $z=0$. Evaluating the residue will give the remaining part of the result.
Update:
On the above, I chose the branch cut of a function of the form $(p-p_{0})^{\alpha}(p-p_{1})^{\beta}$ such that it exists only between the two algebraic branch points. However, we in general ought to have two infinitely long cuts emanating from the branch points. To have a single cut only between the two branch points, the two original branch cuts should be aligned so that they coincide and must somehow cancel each other in the overlapping region. It is possible only when $\alpha+\beta$ is an integer.
To illustrate this point, suppose that $p_{0}, p_{1} \in \mathbb{R}$ and that $p_{0}<p_{1}$. We take the branch cuts to be $[p_{0},\infty)$ and $[p_{1},\infty)$. Then, consider polar representations
\begin{equation}
p-p_{0} = r_{0} e^{i\theta_{0}}, \quad p-p_{1} = r_{1} e^{i\theta_{1}}.
\end{equation}
The angles $\theta_{0}$ and $\theta_{1}$ have discontinuous jumps $\Delta\theta_{0}=\Delta\theta_{1}=2\pi$ at the respective branch cuts, i.e., $[p_{0},\infty)$ and $[p_{1},\infty)$. Now let's consider
\begin{equation}
(p-p_{0})^{\alpha}(p-p_{1})^{\beta} = r_{0}^{\alpha} r_{1}^{\beta} e^{i(\alpha\theta_{0}+\beta\theta_{1})}.
\end{equation}
At the overlapping region, i.e., $[p_{1},\infty)$, the argument of this quantity jumps by
\begin{equation}
\alpha\Delta\theta_{0} + \beta\Delta\theta_{1} = 2\pi (\alpha+\beta).
\end{equation}
Therefore, when $\alpha+\beta$ is an integer, there is no discontinuity in $e^{i(\alpha\theta_{0}+\beta\theta_{1})}$ and hence in $(p-p_{0})^{\alpha}(p-p_{1})^{\beta}$ across $[p_{1},\infty)$. Two overlapping cuts in this region cancel each other and the function is made analytic there.
This kind of cancellation can happen whenever branch cuts from any number of algebraic branch points coincide and the powers add up to an integer.
An infinite series whose sum is $\pi/4G$.
Applying the idea of @reuns in the comment.
Let $\beta(s)=L(s,\chi_4)$ be Dirichlet beta function. Then we have
$$
\sum_{n=1}^{\infty} \frac{\chi_4(n)\phi(n)}{n^{s+1}} = \frac{\beta(s)}{\beta(s+1)}. \ \ \ (1)
$$
Put $s=1$, we obtain the infinite series
$$
\sum_{n=1}^{\infty} \frac{\chi_4(n)\phi(n)}{n^2}=\frac{\beta(1)}{\beta(2)}=\frac{\pi}{4G}.
$$
By definition of $\chi_4$ below, we may also write
$$
\sum_{k=0}^{\infty} \frac{(-1)^k\phi(2k+1)}{(2k+1)^2}=\frac{\pi}{4G}.
$$
Derivation of (1):
$$
\begin{align}\frac{L(s,\chi_4)}{L(s+1,\chi_4)}&= \prod_p \frac{ 1-\frac{\chi_4(p)}{p^{s+1}}}{1-\frac{\chi_4(p)}{p^s}}\\
&=\prod_p \left(1-\frac{\chi_4(p)}{p^{s+1}}\right)\left(1+\frac{\chi_4(p)}{p^s} + \frac{\chi_4^2(p)}{p^{2s}} + \cdots \right)\\
&=\prod_p \left(1+\frac{\chi_4(p)}{p^s}\left(1-\frac1p\right) + \frac{\chi_4^2(p)}{p^{2s}}\left(1-\frac1p\right)+\cdots\right)\\
&=\sum_{n=1}^{\infty} \frac{\chi_4(n) \frac{\phi(n)}n}{n^s} \ \textrm{ if } \ \Re s >1.
\end{align}
$$
This derivation requires the change of order of summation, which is valid under absolute convergence. So, this is okay if $\Re s >1$.
The convergence of (1) at $s=1$.
For $s=1$, the convergence is a bit more subtle. We proceed as follows.
$$
\begin{align}
\sum_{n\leq x} \frac{\chi_4(n)\phi(n)}{n^2} &= \sum_{n\leq x} \frac{\chi_4(n)}n \sum_{d|n} \frac{\mu(d)}d \ \left(\textrm{We have} \ \frac{\phi(n)}n=\sum_{d|n}\frac{\mu(d)}d\right)\\
&=\sum_{d\leq x} \sum_{k\leq \frac xd} \frac{\mu(d)\chi_4(d)\chi_4(k)}{d^2k} \ (\textrm{Here, we use} \ n=dk )\\
&=\sum_{d\leq x} \frac{\mu(d)\chi_4(d)}{d^2} \sum_{k\leq \frac xd} \frac{\chi_4(k)}k\\
&=\sum_{d\leq x}\frac{\mu(d)\chi_4(d)}{d^2} \left( L(1,\chi_4) + O\left(\frac dx\right)\right)\\
&=\frac{L(1,\chi_4)}{L(2,\chi_4)} +O\left(\frac1x\right) + O\left(\sum_{d\leq x} \frac1{dx}\right)\\
&=\frac{L(1,\chi_4)}{L(2,\chi_4)} + O\left(\frac{\log x}x\right) \\
&=\frac{\pi}{4G}+ O\left(\frac{\log x}x\right).
\end{align}
$$
Definitions of $\chi_4, \phi, \mu$.
Here $\chi_4(n)$ (Dirichlet character modulo $4$) and $\phi(n)$ (Euler Totient function), $\mu(n)$ (Mobius function) are defined by
$$
\chi_4(n)=\begin{cases} 0 &\mbox{ if } n \mbox{ is even}\\
(-1)^k &\mbox{ if } n=2k+1\end{cases}
$$
$$
\phi(n)=n \prod_{p|n} \left(1-\frac 1p\right) \mbox{ is the number of $1\leq k\leq n$ coprime to $n$. }
$$
$$
\mu(n)=\begin{cases} 0 &\mbox{ if } n \ \mbox{ is not square-free}\\
(-1)^{\omega(n)} &\mbox{ if } n \ \mbox{is square-free, }\end{cases}
$$
where $\omega(n)$ is the number of distinct prime factors of $n$.
Best Answer
The answer is yes. The method is detailed in the paper by S. K. Lucas. The approximate fractions are obtained from truncating the exact continued fraction of the respected numbers, so the signs are alternating. We first give a few examples.
Results
For $\pi$
The continued fractions are $3, 22/7, 333/106, 355/113, 103993/33102, \dots$.
\begin{align} \pi - \frac{333}{106} &= \int_0^1 \frac{x^4 \, (1-x)^5 \, \left(74 \, x^2-53 \, x+21\right)}{106 \left(x^2+1\right)} \, dx. \\ \frac{355}{113} - \pi &= \int_0^1 \frac{x^{10} \, (1-x)^8 \, \left(886+95\,x^2\right)}{3164 \left(x^2+1\right)} \, dx. \end{align}
For $\pi^2$
The truncated continued fractions are $9, 10, 69/7, 79/8, 227/23, 10748/1089, \dots$,
\begin{align} \pi^2-\frac{69}{7} &= \int_0^1 \frac{4 \, x^{4} \, (1 - x)^3 \left(64 x^2 -39 x + 25\right)} {13 \, (1 + x^2) } \log(x^{-1}) \, dx \\ &= \int_0^1 \frac{24 \, x^{6} \, (1 - x)^2 \left(119 - 72 \, x^2\right)} {191 \, (1 + x^2) } \log(x^{-1}) \, dx. \end{align}
\begin{align} \frac{79}{8} - \pi^2 &= \int_0^1 \frac{4 \, x^6 \, (1-x)^3 (49 - 51 x + 100 x^2)} {17 \, (1 + x^2) } \log(x^{-1}) \, dx \\ &= \int_0^1 \frac{4 \, x^3 \, (1-x)^4 (25 + 2254 x^2)} {743 \, (1 + x^2) } \log(x^{-1}) \, dx \\ &= \int_0^1 \frac{ 24 \, x^5 \, (1-x)^2 \left[37 \, (x^2 + 1) - 73 \, x\right] } { 73 \, (1 + x^2) } \log(x^{-1}) \, dx \end{align}
\begin{align} \pi^2-\frac{227}{23} &= \int_0^1 \frac{4 \, x^{19} \, (1 - x)^4 \left(61847 x^2+87524\right)} {8559 \, (1 + x^2) } \log(x^{-1}) \, dx. \end{align}
For $\pi^3$
The truncated continued fractions are $31, 4930/159, 14821/478, \dots$.
\begin{align} \pi^3-31 &= \int_0^1 \frac{8 \, x^5 \, (1-x)^2 \, \left(324889-120736 \, x^2\right)} {445625 \, (1 + x^2) } \log^2 x \, dx\\ \frac{4930}{159}-\pi^3 &= \int_0^1 \frac{4 \, x^{10} \, (1-x)^4 \, \left(695774836+470936528857 \, x^2\right)} {470240754021 \, (1 + x^2) } \log^2 x \, dx. \end{align}
For $\pi^4$
The truncated continued fractions are $97, 195/2, 487/5, 1656/17, 2143/22, \dots$.
\begin{align} \pi^4-97 &= \int_0^1 \frac{240 \, x^{4} \, (1 - x)^{2} \,\left(3522267 x^2+1681375\right) } {3221561 \, (1 + x^2) } \log^3(x^{-1}) \, dx \\ \frac{195}{2}-\pi^4 &= \int_0^1 \frac{192 \, x^{6} \, (1 - x)^{2} \, \left(5657688 x^2+3056473\right) } {3641701 \, (1 + x^2) } \log^3(x^{-1}) \, dx. \\ \pi^4-\frac{487}{5} &= \int_0^1 \frac{15 \, x^{8} \, (1 - x)^{2} \, \left(3293858975 x^2+746556831\right) } {278611172 \, (1 + x^2) } \log^3(x^{-1}) \, dx. \\ \frac{1656}{17}-\pi^4 &= \int_0^1 \frac{480 \, x^{7} \, (1 - x)^{4} \, \left(8555775811 x^2+2883779820\right) } {39703971937 \, (1 + x^2) } \log^3(x^{-1}) \, dx. \\ \pi^4-\frac{2143}{22} &= \int_0^1 \frac{480 \, x^{31} \, (1 - x)^{4} \, \left(4071997316165706379 x^2+175446796437023645180\right) } {1199623593846005571607 \, (1 + x^2) } \log^3(x^{-1}) \, dx. \end{align}
Method
The idea is simple. We basically combine the following identities.
(1) \begin{align} \int_0^1 \log^{s-1}\left( x^{-1} \right) x^k \, dx = \frac{(s-1)!}{(k+1)^s}. \end{align}
For an even $s$
(2a) \begin{align} \int_0^1 \frac{ \log^{s-1}(x^{-1}) \, x }{1+x^2} \, dx &=2^{-s} \int_0^\infty \frac{ t^{s-1} }{ e^t + 1 } \, dt \\ &=2^{-s} \int_0^\infty t^{s-1} \left( e^{-t} - e^{-2\,t} + e^{-3t} - \cdots \right) \, dt \\ &= \frac{ (s-1)! \, (2^s - 2)}{4^s} \zeta(s) \\ &= \frac{ (1-2^{1-s}) \, |B_s| }{2 \, s} \, \pi^s. \end{align} where $\zeta(s)$ is the Riemann zeta function, $B_s$ is the Bernoulli number, which is rational. The last step is well known.
Similarly, for an odd $s$,
(2b) \begin{align} \int_0^1 \frac{ \log^{s-1}(x^{-1}) }{1+x^2} \, dx &=\int_0^\infty \frac{ t^{s-1} \, e^{-t} }{ e^{-2t} + 1 } \, dt \\ &=\int_0^\infty t^{s-1} \left( e^{-t} - e^{-3\,t} + e^{-5\,t} -\cdots \right) \, dt \\ &= (s-1)! \, \left(1-\frac{1}{3^s}+\frac{1}{5^s}-\cdots \right) \\ &= (s-1)! \, \beta(s) = \frac{|E_{s-1}|}{2^{s+1}} \, \pi^s, \end{align} where $E_s$ is the Euler number, which is also rational.
This means \begin{align} \frac{\pi}{4} &= \int_0^1 \frac{ 1 }{1+x^2} \, dx, \\ \frac{\pi^2}{48} &= \int_0^1 \frac{\log\left( x^{-1} \right) x}{1+x^2} \, dx\\ \frac{\pi^3}{16} &= \int_0^1 \frac{\log^2\left( x^{-1} \right) }{1+x^2} \, dx\\ \frac{7\,\pi^4}{192} &= \int_0^1 \frac{\log^3\left( x^{-1} \right) x}{1+x^2} \, dx. \end{align}
Now suppose we have a polynomial $P(x) = Q(x)(1 + x^2) + R(x)$, and we want $$ \int_0^1 \frac{ \log^s(x^{-1}) \, P(x) } { 1 + x^2 } \, dx = \pi^s - A, $$ where $A$ is an approximation of $\pi^s$ (we have assumed the possible sign, the case of negative sign is similar). To satisfy this equation, we demand,
\begin{align} \int_0^1 \frac{ \log^s(x^{-1}) \, R(x) } { 1 + x^2 } \, dx &= \pi^s \\ \int_0^1 \log^s(x^{-1}) \, Q(x) \, dx &= - A, \end{align}
This requires \begin{align} R(x) &= \begin{cases} \dfrac{ 2^{s+1} } { |E_{s-1}|} & \mathrm{for\; odd\;} s \\ \dfrac{ 2 \, s } { (1-2^{1-s}) \, |B_s| } x & \mathrm{for\; even\;} s \end{cases} \\ \sum_{n=0} \frac{ (s-1)! }{(n+1)^s} q_n &= -A, \end{align} where $Q(x) = \sum_{n=0} q_n \, x^n$.
Using these to rules to design $P(x)$, we get the above formulas. Particularly, we studied the form \begin{align} P(x) = x^u \,(1-x)^v (a \, x^2 + b \, x + c). \end{align}
For a particular set of $u$ and $v$, the constraint for $R(x)$ determines two parameters, say $b$ and $c$. The constraint for $A$ determines $a$. We then check if $a \, x^2 + b \, x + c$ is nonnegative definite. We then vary $u$ and $v$ to seek a simple combination of $a$, $b$ and $c$.