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 $c$.
