$\newcommand\HGF{_2\!\tilde{F}_1}$Mathematica is able to compute these limits, the result is in terms of a partial derivative of the regularized hypergeometric function $\HGF$. For $k=2$ I find
$$\lim_{k\rightarrow 2}x^{-k}I_k(x)=\frac{11}{3}-\gamma_{\text{Euler}}+\ln x-\frac{1}{6x^3}\left({\HGF^{(0,0,1,0)}}(-3,1,-2;x)+{\HGF^{(1,0,0,0)}}(-3,1,-2;x)\right).$$
The superscript notation is Mathematica's way of indicating which variable to differentiate. As a test, for $x=1/2$ the right-hand-side evaluates to $-4.833333\cdots=-29/6$, which agrees with a numerical evaluation of $\lim_{k\rightarrow 2}[\beta_x( -1 - k, 0) + H_{-2 - k}]$.

The corresponding formula for integer $k\geq 2$ is
$$\lim_{n\rightarrow k}x^{-n}I_n(x)=c_k-\gamma_{\text{Euler}}+\ln x-\frac{1}{(k+1)!x^{k+1}}\left({\HGF^{(0,0,1,0)}}(-k-1,1,-k;x)+ {\HGF^{(1,0,0,0)}}(-k-1,1,-k;x)\right).$$
The fraction $c_k$ is twice the constant term in an expansion of $H_{-2-n}$ around $n=k$.
I don't have a closed-form expression for $c_k$, the first few values are $11/3, 25/6, 137/30, 49/10, 363/70$, for $k=2,3,4,5,6$. 

**Update:** As pointed out by Peter Taylor: $c_k=2H_{k+1}$.

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Mathematica code.

     k=6;
     Series[Gamma[a]x^a Hypergeometric2F1Regularized[a,1-b,a+1,x],{a,-1-k,0}]/.b->0//Normal// FullSimplify;
     %/.a->-1-x;
     Series[HarmonicNumber[-2-x],{x,k,0}]//Normal;
     %+%%//FullSimplify

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The asympotics near $x=0$ and $x=1$ is
$$\lim_{n\rightarrow k}x^{-n}I_n(x)\rightarrow\begin{cases}
-\frac{1}{(k+1)x^{k+1}}&\text{for}\;\;x\rightarrow 0\\
 -\ln(1-x)&\text{for}\;\;x\rightarrow 1
\end{cases}$$

Here is a plot of $\lim_{n\rightarrow k}x^{-n}I_n(x)$ for $k=2$.

<IMG SRC="https://ilorentz.org/beenakker/MO/betaharmonic_1.png" WIDTH="400"/>

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**Special cases:**


$$k=2:\quad\lim_{k\rightarrow 2}x^{-k}I_k(x)=\frac{(x-1) (11 x^2+5 x+2)}{6 x^3}-2 \,\text{arctanh}\,(1-2 x)$$ 

$$k=3:\quad\lim_{k\rightarrow 2}x^{-k}I_k(x)=\frac{(x-1) (25 x^3+13 x^2+7 x+3)}{12 x^4}-2 \,\text{arctanh}\,(1-2 x)$$

$$k=4:\quad\lim_{k\rightarrow 2}x^{-k}I_k(x)=\frac{(x-1) (12 + 27 x + 47 x^2 + 77 x^3 + 137 x^4)}{60 x^5}-2 \,\text{arctanh}\,(1-2 x)$$

$$k=5:\quad\lim_{k\rightarrow 2}x^{-k}I_k(x)=\frac{(x-1)(10 + 22 x + 37 x^2 + 57 x^3 + 87 x^4 + 147 x^5)}{60 x^6}-2 \,\text{arctanh}\,(1-2 x)$$