You can use the identity given by the Euler Beta function
$$\int_{0}^{1}x^{a-1} (1-x)^{b-1} \,dx=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$
to state:
$$S=\sum_{k=1}^{+\infty}\frac{(-1)^{k+1}}{k!}\Gamma(k/2)^2=\sum_{k=1}^{+\infty}\frac{(-1)^{k-1}}{k}\int_{0}^{1}\left(x(1-x)\right)^{k/2-1}\,dx $$
and by switching the series and the integral:
$$ S = \int_{0}^{1}\frac{\log(1+\sqrt{x(1-x)})}{x(1-x)}dx = 2\int_{0}^{1/2}\frac{\log(1+\sqrt{x(1-x)})}{x(1-x)}dx,$$
$$ S = 2\int_{0}^{1/2}\frac{\log(1+\sqrt{1/4-x^2})}{1/4-x^2}dx = 4\int_{0}^{1}\frac{\log(1+\frac{1}{2}\sqrt{1-x^2})}{1-x^2}dx,$$
$$ S = 4\int_{0}^{\pi/2}\frac{\log(1+\frac{1}{2}\sin\theta)}{\sin\theta}d\theta.$$
Now Mathematica gives me $\frac{5\pi^2}{18}$ as an explicit value for the last integral, but probably we are on the wrong path, and we only need to exploit the identity
$$\sum_{k=1}^{+\infty}\frac{1}{k^2\binom{2k}{k}}=\frac{\pi^2}{18}$$
that follows from the Euler acceleration technique applied to the $\zeta(2)$-series. The other "piece" (the $U$-piece in the Marty Cohen's answer) is simply given by the Taylor series of $\arcsin(z)^2$. More details to come.
As a matter of fact, both approaches lead to an answer.
The (Taylor) series approach, as Bhenni Benghorbal shows below, leads to the identity:
$$\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k!}\Gamma^2\left(\frac{k}{2}\right)x^k= 2 \arcsin \left( x/2 \right) \left(\pi - \arcsin \left( x/2\right) \right),\tag{1}$$
while the integral approach, as Achille Hui pointed out in the comments, leads to:
$$\begin{eqnarray*}\int_{0}^{\pi/2}\frac{\log(1+\frac{1}{2}\sin\theta)}{\sin\theta}\,d\theta&=&\int_{0}^{1}\log\left(1+\frac{t}{1+t^2}\right)\frac{dt}{t}\\&=&\int_{0}^{1}\frac{\log(1-t^3)-\log(1-t)-\log(1+t^2)}{t}\,dt\\&=&\int_{0}^{1}\frac{-\frac{2}{3}\log(1-t)-\frac{1}{2}\log(1+t)}{t}\,dt\\&=&\frac{1}{6}\sum_{k=1}^{+\infty}\frac{4+3(-1)^k}{k^2}=\frac{1}{6}\left(4-\frac{3}{2}\right)\zeta(2)=\frac{5\pi^2}{72}.\end{eqnarray*}\tag{2}$$
Thanks to both since now this answer may become a reference both for integral-log-ish problems like $(2)$ and for $\Gamma^2$-series like $(1)$.
Update 14-06-2016. I just discovered that this problem can also be solved by computing
$$ \int_{-1}^{1} x^n\, P_n(x)\,dx, $$
where $P_n$ is a Legendre polynomial, through Bonnet's recursion formula or Rodrigues' formula. Really interesting.
In order to approximate the solution, I built for $\binom{1}{x}$ the
$$\binom{1}{x}\sim \frac{4}{\pi }+\left(\frac{16}{\pi }-2 \pi \right)
\left(x-\frac{1}{2}\right)^2+\left(110-\frac{576}{\pi }+24 \pi -\frac{\pi
^2}{6}\right) \left(x-\frac{1}{2}\right)^4+$$ $$\left(-608+\frac{2816}{\pi }-96 \pi
+\frac{4 \pi ^2}{3}\right)
\left(x-\frac{1}{2}\right)^6+\left(928-\frac{4096}{\pi }+128 \pi -\frac{8 \pi
^2}{3}\right)
\left(x-\frac{1}{2}\right)^8$$ which exactly matches the function, first and second derivatives values for $x=0,\frac12,1$.
Now, concerning the integrals
$$I_n=\int_0^1 \left(x-\frac{1}{2}\right)^{2n} \,\,\,\frac{\log^2(1-x)}{x}\, dx$$ we have
$$I_0=2 \zeta (3)\qquad I_1=\frac{\zeta (3)}{2}-\frac{1}{4}\qquad I_2=\frac{\zeta (3)}{8}-\frac{71}{864}$$ $$I_3=\frac{\zeta (3)}{32}-\frac{10051}{432000}\qquad I_4=\frac{\zeta (3)}{128}-\frac{116069}{18522000}$$
This finally leads to
$$\int_0^1 \binom{1}{x}\frac{\log^2(1-x)}{x}\, dx\sim 2 \zeta (3)-\frac{13129439}{18522000}+\frac{4032568}{1157625 \pi }-\frac{94427
\pi }{2315250}-\frac{45551 \pi ^2}{74088000}$$ which is numerically $2.6698874395$ to be compared to the "exact" $2.6698874550$ given by @dust05 in comments.
We could do much better using a few more terms for the approximation.
Edit
Instead of using the built approximation given above, I used the Taylor expansion of $\binom{1}{x}$ (you can get it using Wolfram Alpha; the problem I faced is that the odd terms are not recognized to be $0$ and I do not see how to tansform the digamma terms into simpler expressions).
So, I computed
$$I_k=\sum_{n=0}^k c_n \int_0^1 \left(x-\frac{1}{2}\right)^{2n} \,\,\,\frac{\log^2(1-x)}{x}\, dx$$ As a function of $k$, the numerical results are reported below
$$\left(
\begin{array}{cc}
k & I_k \\
0 & 3.06101276824936274 \\
1 & \color{red} {2.6}4320918239538048 \\
2 & \color{red} {2.6}7090489976321907 \\
3 & \color{red} {2.6698}6268034682524 \\
4 & \color{red} {2.669887}87367277585 \\
5 & \color{red} {2.6698874}4975680354 \\
6 & \color{red} {2.66988745}500671797 \\
7 & \color{red} {2.66988745495}686919 \\
8 & \color{red} {2.66988745495724}361 \\
9 & \color{red} {2.6698874549572413}3 \\
10 & \color{red} {2.66988745495724134}
\end{array}
\right)$$
Taking into account @Mariusz Iwaniuk 's hint, that is to say
$$\binom{1}{x}=\frac{\sin (\pi x)}{\pi (1-x) x}$$ the coefficients $c_n$ can write
$$c_n=i\,\frac{ 2^{2 n+1} }{\pi (2 n)!}\left(\Gamma \left(2 n+1,\frac{i \pi }{2}\right)-\Gamma \left(2 n+1,-\frac{i \pi }{2}\right)\right)$$
Concerning the integrals
$$J_n=\int_0^1 \left(x-\frac{1}{2}\right)^{2n} \,\,\,\frac{\log^2(1-x)}{x}\, dx$$ they write
$$J_n=2^{1-2 n} \zeta (3)-\frac {a_n} {b_n}$$
$$\left(
\begin{array}{ccc}
0 & 0 & 1 \\
1 & 1 & 4 \\
2 & 71 & 864 \\
3 & 10051 & 432000 \\
4 & 116069 & 18522000 \\
5 & 52752017 & 32006016000 \\
6 & 145759321889 & 340800058368000 \\
7 & 329587937534753 & 2994950912937984000 \\
8 & 42159304836511 & 1497475456468992000 \\
9 & 844375600417012397 & 117713550682114523136000 \\
10 & 11769137630214586888219 & 6459177953028988113518592000 \\
11 & 11930404954629448855339 & 25836711812115952454074368000 \\
12 & 18359838608628619185581941 & 157177636309007396754361417728000 \\
13 & 9275258078308733536880688959 & 314355272618014793508722835456000000
\end{array}
\right)$$
Computed up to $n=50$, there is very good correlation $(R^2=0.999994)$
$$\log \left(\frac{a_n}{b_n}\right)=\alpha +\beta \,n$$
$$\begin{array}{clclclclc}
\text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\
\alpha & +0.507370 & 0.029184 & \{+0.448659,+0.566081\} \\
\beta & -1.378775 & 0.000996 & \{-1.380779,-1.376772\} \\
\end{array}$$
Update
Concerning
$$\int_0^1 \binom{1}{x}\frac{\log(1-x)}{x}\,dx$$ I used my initial approximation and ended with
$$\frac{8219}{9800}-\frac{42704}{11025 \pi }+\frac{428 \pi }{11025}-\frac{175933
\pi ^2}{1058400} \approx -1.9128812062$$ while numerical integration gives $-1.9128812187$.
Best Answer
$$S_n=\sum_{r=0}^{n}\frac{(-1)^r\binom{n}{r}\log(n+r+1) }{n+r+1}=-\frac{\partial }{\partial s}\sum_{r=0}^{n}\frac{(-1)^r\binom{n}{r}}{(n+r+1)^s}\,\bigg|_{s=1}$$ Using $\,\displaystyle \frac{1}{(n+r+1)^s}=\frac1{\Gamma(s)}\int_0^\infty t^{s-1}e^{-t(n+r+1)}dt$ $$S_n=-\frac{\partial }{\partial s}\,\bigg|_{s=1}\frac1{\Gamma(s)}\int_0^\infty t^{s-1}e^{-t(n+1)}\left(\sum_{r=0}^n(-1)^r\binom{n}{r}e^{-rt}\right)dt$$ $$=-\frac{\partial }{\partial s}\,\bigg|_{s=1}\frac1{\Gamma(s)}\int_0^\infty t^{s-1}e^{-t(n+1)}(1-e^{-t})^ndt$$ $$\overset{x=e^{-t}}{=}-\gamma\int_0^1x^n(1-x)^ndx-\int_0^1\ln(-\ln x)x^n(1-x)^ndx$$ $$=-\gamma \,B(n+1;n+1)-\int_0^1\ln(-\ln x)x^n(1-x)^ndx$$ I'm not sure whether the last integral has a nice closed form, though its asymptotics at $n\to\infty$ is straightforward: $$-\int_0^1\ln\Big(\ln \frac1x\Big)x^n(1-x)^ndx=-\int_{-1/2}^{1/2}\ln\Big(\ln \frac2{2x+1}\Big)\big(\frac12+x\big)^n\big(\frac12-x\big)^ndx$$ $$\overset{x=\frac t2}{=}-\frac1{2\cdot4^n}\int_{-1}^1\ln\Big(\ln \frac2{t+1}\Big)(1-t^2)^ndt\sim-\,\frac{\ln(\ln2)}{2\cdot4^n}\sqrt\frac\pi n$$ where the Laplace' method was used in the last line.
Using the same approach we find that $$-\gamma \,B(n+1;n+1)\sim-\,\frac{\gamma}{2\cdot4^n}\sqrt\frac\pi n\,\, \text{at}\,\, n\to\infty$$ and the desired asymptotics $$\boxed{\,\,S_n\sim-\,\frac{\gamma+\ln(\ln2)}{2\cdot4^n}\sqrt\frac\pi n\,\,\, \text{at}\,\, n\to\infty\,\,}$$ The numeric check with WolframAlpha confirms the result.