Let $N$ be a positive integer and $c_1,\ldots,c_N$ non-negative real numbers. Denote $f(x)=((x+c_1)\ldots (x+c_N))^{-1}$.

**Lemma 1**. For all integer $d\geqslant 0$ and all $x>0$ we have $(-1)^df^{(d)}(x)>0$.

**Proof.** Induction in $d$. Base $d=0$ is trivial. For doing an induction step it suffices to note that $-f'$ is a sum of $N$ functions of the same type.

**Lemma 2.** For a non-negative integer $d$, real $a$, and a continuous real function $f(x)$ defined on the segment $[a,a+d]$ and having $d$ derivatives on the interval $(a,a+d)$ ($d$-th derivative may be discontinuous) there exists $\theta\in (0,d)$ such that $$f^{(d)}(a+\theta)=\sum_{k=0}^d (-1)^{d-k}{d\choose k}f(a+k)=:c$$

**Proof.** Denote $h(x)=\sum_{k=0}^d f(a+k){x\choose k}{d-x\choose d-k}$.Then $h(x)=\frac{c}{d!}x^d+\ldots$ is a polynomial of degree at most $d$ and $h(k)=f(a+k)$ for all $k=0,1,\ldots,d$. Therefore, by Rolle's theorem, the function $g(x):=f(a+x)-h(x)$, which has roots at points $0,1,\ldots,d$, satisfies $g^{(d)}(\theta)=0$ for certain $\theta\in (0,d)$. This $\theta$ is what we need.

Now call a function $f(k)$ good if it has a type $f(k)=\frac{C}{(k+r)(k+r+1)\ldots (k+s)}$ for certain $C>0$ and positive integers $r\leqslant s$.
These two lemmata already yield that when $q+1\leqslant 2t$, then $\sum_{k=0}^{2 t} (-1)^k{q+1\choose k}f(k) \geqslant 0$ that is your claim (is it?)

For larger values of $q$, we induct on $q$ with use of Pascal's recurrence ${q+1\choose k}={q\choose k}+{q\choose k-1}$ which allows to write
\begin{align*}
\sum_{k=0}^{2t} (-1)^k {q+1\choose k}f(k)
&=\sum_{k=0}^{2t} (-1)^k {q\choose k}f(k)-\sum_{k=0}^{2t-1} (-1)^k{q\choose k}f(k+1) \\
&={q\choose 2t}f(2t+1)+\sum_{k=0}^{2t}(-1)^k{q\choose k}(f(k)-f(k+1)).
\end{align*}
Since the function $k\to f(k)-f(k+1)$ is good, we reduced $q+1$ to $q$.