First we derive the formulas, afterwards we study some applications of these.
1) Integral representation of the unified harmonic sum
$$U(\sigma ,n,p)=U_{n,p}(\sigma)=\sum _{k=1}^n \frac{\sigma ^k}{k^p}
\\
=\sum _{k=1}^n \sigma ^k \left(\frac{1}{\Gamma (p)}\int _0^{\infty }t^{p-1}\exp \left(-k t\right)\,dt\right)
\\
=\frac{1}{\Gamma (p)}\int _0^{\infty }t^{p-1}\left(\sum _{k=1}^n \sigma ^k \exp \left(-k t\right)\right)\,dt
$$
hence
$$U(\sigma ,n,p)=U_{n,p}(\sigma)=\frac{1}{\Gamma (p)}\int_0^{\infty } t^{p-1}\sigma e^{-t}\frac{ 1-\left(\sigma e^{-t}\right)^n}{1-\sigma e^{-t}} \, dt\tag{s1}$$
After the transformation $t\to \log(\frac{1}{x})$ this form of the integral is obtained
$$U(\sigma ,n,p)=U_{n,p}(\sigma)=\frac{\sigma }{\Gamma (p)} \int_0^1 \log ^{p-1}\left(\frac{1}{x}\right)\frac{1-(\sigma x)^n}{1-\sigma x} \, dx\tag{s1a}$$
For $p=1$ this simplifies to
$$U(\sigma,n) = \sum _{k=1}^n \frac{\sigma ^k}{k}=\sigma \int_0^1 \frac{1-(\sigma x)^n}{1-\sigma x} \, dx=\int_0^{\sigma} \frac{1-y^n}{1-y} \, dy\tag{s1b}$$
Here we have abbreviated $U(\sigma,n,p=1) = U(\sigma,n)$.
2) Representation of the unified harmonic sum as an infinite sum
A power series expansion of the denominator in $(s1)$ gives
$$U(\sigma ,n,p)=U_{n,p}(\sigma)=\frac{1}{\Gamma (p)}\int_0^{\infty } t^{p-1}\sigma e^{-t}\frac{ 1-\left(\sigma e^{-t}\right)^n}{1-\sigma e^{-t}} \, dt
\\
= \frac{1}{\Gamma (p)} \int_0^{\infty } t^{p-1}
\left( 1-\left(\sigma e^{-t}\right)^n\right) \sum_{m=1}^{\infty}(\sigma e^{-t})^m \, dt
\\
= \frac{1}{\Gamma (p)}\sum_{m=1}^{\infty} \int_0^{\infty } t^{p-1}
\left( 1-\left(\sigma e^{-t}\right)^n\right) (\sigma e^{-t})^m \, dt
\\
= \frac{1}{\Gamma (p)}\sum_{m=1}^{\infty} \int_0^{\infty } \left(
t^{p-1}(\sigma e^{-t})^m - t^{p-1}\left(\sigma e^{-t}\right)^{m+n}\right)\, dt $$
The two integrals can be easily done so that we get
$$U(\sigma ,n,p)=U_{n,p}(\sigma)= \sum _{m=1}^{\infty } \left(\frac{\sigma ^m}{m^p}-\frac{\sigma ^{m+n}}{(m+n)^p}\right)\tag{s2}$$
Notice that this relation can also be written in terms of standard functions as
$$U(\sigma,n,p)= \operatorname{Li}_p(\sigma )-\sigma ^{n+1} \Phi (\sigma ,p,n+1)\tag{s2a}$$
where $\operatorname{Li}_p(z)=\sum_{k=1}^{\infty}\frac{z^k}{k^p} $ is the polylog function and $\Phi (z ,p,c)= \sum_{k=0}^{\infty}\frac{z^k}{(k+c)^p}$ is the Lerch phi function.
3) The generating function of the unified harmonic sum
The generating function of order $q$ is defined as
$$g_{p,q}(\sigma,z )=\sum_{n=1}^{\infty} \frac{z^n}{n^q}U_{n,p}(\sigma)$$
Inserting $U$ from ${s1}$ we have
$$g_{p,q}(\sigma,z )=
=\sum_{n=1}^{\infty} \frac{z^n}{n^q}\left(\frac{1}{\Gamma (p)}\int_0^{\infty } t^{p-1}\sigma e^{-t}\frac{ 1-\left(\sigma e^{-t}\right)^n}{1-\sigma e^{-t}}\right) \, dt
\\
=\frac{1}{\Gamma (p)}\int_0^{\infty } t^{p-1}\sigma e^{-t}\frac{1}{1-\sigma e^{-t}}\sum_{n=1}^{\infty} \frac{z^n}{n^q}\left( 1-\left(\sigma e^{-t}\right)^n\right) \, dt
$$
The sums can be expressed by polylog functions defined as
$$\operatorname{Li}_{q}(z)=\sum_{n=1}^{\infty} \frac{z^n}{n^q} $$
so that we get finally
$$g_{p,q}(\sigma,z )=\frac{1}{\Gamma (p)}\int_0^{\infty } t^{p-1}(\sigma e^{-t})\frac{ \operatorname{Li}_q(z)-\operatorname{Li}_q\left(e^{-t} z \sigma \right)}{1-\sigma e^{-t} } \, dt\tag{s3}$$
We can get a simplified version doing the integral of the first term giving
$$g_{p,q}(\sigma,z )= \operatorname{Li}_p(\sigma ) \operatorname{Li}_q(z)-\frac{1}{\Gamma (p)}\int_0^{\infty } \frac{t^{p-1} \left(\sigma e^{-t}\right) \operatorname{Li}_q\left(e^{-t} z \sigma \right)}{1-\sigma e^{-t}} \, dt\tag{s4}$$
An alternative representation as a double integral can be obtained replacing $\frac{1}{n^q}$ with an integral, resulting in
$$g_{p,q}(\sigma,z )=\frac{1}{\Gamma (p) \Gamma (q)}\int_0^\infty \int_0^\infty\,ds\,dt \frac{ t^{p-1} s^{q-1} e^{-t-s} (\sigma z) }{ \left(1-e^{-s} z\right) \left(1-\sigma z e^{-s-t}\right)}\tag{s5}$$
If we now do the $t$-integral we get a simpler alternative to $(s3)$
$$g_{p,q}(\sigma,z )= \frac{1}{\Gamma(q)} \int_0^\infty s^{q-1} \frac{\operatorname{Li}_p\left(e^{-s} z \sigma \right)}{1-e^{-s} z}\,ds\tag{s6}$$
4) Micellaneous
The following symmetry relations for even an odd indices hold
$$U_{2n,p}(\sigma) = - U_{2n,p}(-\sigma) +U_{n,p}(\sigma^2) \tag{s4.1}$$
$$U_{2n+1,p}(\sigma) = - U_{2n+1,p}(-\sigma) +U_{n,p}\sigma^2)\tag{s4.2} $$
For $\sigma = 1$ these reduce to
$$H_{2n,p} = \overline{H}_{2n,p} +H_{n,p} \tag{s4.1a}$$
$$H_{2n+1,p} = \overline{H}_{2n+1,p} +H_{n,p}\tag{s4.2b} $$
By repeated partial integration of ${(s1b)}$ we can easily derive the following asymptotic expansion valid for $\sigma \ne 1$
$$U(\sigma,n) = \sigma \int_0^1 \frac{1-(\sigma x)^n}{1-\sigma x} \, dx \overset{n\to\infty}
\simeq -\log (1-\sigma )
\\-\sigma ^{n+1}
\left(\frac{1}{n (1-\sigma )}-\frac{1}{n^2 (1-\sigma )^2}+\frac{\sigma +1}{n^3 (1-\sigma )^3} \\ -\frac{\sigma ^2+4 \sigma +1}{n^4 (1-\sigma )^4}+\frac{\sigma ^3+11 \sigma ^2+11 \sigma +1}{n^5 (1-\sigma)^5}+O(\frac{1}{n^6})\right)\tag{s4.3}$$
Notice that for $\sigma = 1$ the terms with $\frac{1}{n^3}$ and $\frac{1}{n^5}$ vanish and we recover the asymptotics of $(-\overline{H}_n)$ provided here How to prove the asymptotic expansion $\overline{H}_n \sim \log(2) -(-1)^n\left (\frac{1}{2n}-\frac{1}{4 n^2} +\frac{1}{8n^4}\mp\ldots\right)$?
Recently I became aware of a paper of 2004 https://www.sciencedirect.com/science/article/pii/S0022247X04003920?via%3Dihub (quoted in https://en.wikipedia.org/wiki/Lerch_zeta_function#Asymptotic_expansion) which provided the following expression for the asymptotics of the Lerch Psi function
$$\Phi(z,s,a) \underset{a->\infty} \simeq \frac{1}{1-z} \frac{1}{a^s} + \sum_{k=1}^{N}\frac{ (-1)^k}{k!} \operatorname{Li}_{-k}(z) \frac{(s)_k}{a^{k+s}}+R\tag{s4.4}$$
Using this result the asymptotics of the unified harmonic sum becomes
$$U(\sigma,n) \overset{n\to \infty}\simeq -\log (1-\sigma ) -\sigma ^n \left(\sum _{k=0}^{N} \frac{(-1)^r \operatorname{Li}_{-k}(\sigma )}{n^{k+1}}\right)+O(n^{-N-2})\tag{s4.4a}$$
Notice that the polylog of negative integer index is a quotient of two polynomials. The first few are
$$\left\{\frac{\sigma }{(1-\sigma )},\frac{\sigma }{(1-\sigma )^2},\frac{\sigma ^2+\sigma }{(1-\sigma )^3},\frac{\sigma ^3+4 \sigma ^2+\sigma }{(1-\sigma )^4}\right\}$$
