This is false. A condition like
$$\exists C \in \mathbf{R}_+,~\exists D \in L^q(\mathbf{P}),~\forall n \in \mathbf{N},~\forall z \in \mathbf{R}_+^*, \mathbf{P}[|D_n|>z] \le C \mathbf{P}[|D|>z]$$
would force $(D_n)_{n \in \mathbf{N}}$ to be bounded in $L^q(\mathbf{P})$ since
$$\mathbf{E}[|D_n|^q] = \int_0^\infty \mathbf{P}[|D_n|>z]~qz^{q-1} \mathrm{d}z \le C\int_0^\infty \mathbf{P}[|D|>z]~qz^{q-1} \mathrm{d}z = C\mathbf{E}[|D|^q] .$$ This is not always the case, for ewample when the random variables $D_n/n$ are i.i.d. and not almost surely null.
I complete the answer below: assuming that $(D_n)_{n \in \mathbf{N}}$ to be bounded in $L^q(\mathbf{P})$ is not sufficient. Actually, we have the implications
\begin{eqnarray}
\text{ for some }p>q, (D_n)_{n \in \mathbf{N}} \text{ bounded in } L^q(\mathbf{P})
&\implies& \exists C \in \mathbf{R}_+,~\exists D \in L^q(\mathbf{P}),~\forall n \in \mathbf{N},~\forall z \in \mathbf{R}_+^*, \mathbf{P}[|D_n|>z] \le C \mathbf{P}[|D|>z] \\
&\implies& (|D_n|^q)_{n \in \mathbf{N}} \text{ uniformly integrable }
\\
&\implies& (D_n)_{n \in \mathbf{N}} \text{ bounded in } L^q(\mathbf{P})
\end{eqnarray}
Proof of the first implication
If for some $p>q$, $(D_n)_{n \in \mathbf{N}}$ is bounded in $L^q(\mathbf{P})$, then $C:= \sup\{\mathbf{E}[|D_n|^p] : n \in \mathbf{N}\}$ is finite. For every $z \in \mathbf{R_+}^*$,
$$\mathbf{P}[|D_n|>z] \le \min(1,|z|^{-q})\mathbf{E}[|D_n|^p]$$
so
$$\mathbf{P}[|D_n|>z] \le \min(1,C|z|^{-q})$$
Let $D$ any positive random variable such that for every $z \in \mathbf{R_+}^*$, $\mathbf{P}[D>z] \le \min(1,C|z|^{-q})$. Then
$D \in L^q(\mathbf{P})$ since
$$\mathbf{E}[D^q] = \int_0^{+\infty} \mathbf{P}[D>z]~qz^{q-1} \mathrm{d}z \le \int_0^{+\infty} \min(1,C|z|^{-q})~qz^{q-1}~\mathrm{d}z < +\infty.$$
Proof of the second implication
Assume that we have $C \in \mathbf{R}_+$ and $D \in L^q(\mathbf{P})$ such that
$$\forall n \in \mathbf{N},~\forall z \in \mathbf{R}_+^*, \mathbf{P}[|D_n|>z] \le C \mathbf{P}[|D|>z].$$
Then for every $x \in \mathbf{R_+}^*$
\begin{eqnarray}
\mathbf{E}[|D_n|^q \mathbf{1}_{[|D_n|^q>x]}]
&=& \mathbf{E}\Big[\int_0^\infty \mathbf{1}_{[|D_n|^q>x]}] \mathbf{1}_{[|D_n|>z]}~qz^{q-1} \mathrm{d}z \Big] \\
&=& \mathbf{E}\Big[\int_0^\infty \mathbf{1}_{[|D_n|>\max(x^{1/q},z)]}~qz^{q-1} \mathrm{d}z \Big] \\
&=& \int_0^\infty \mathbf{P}[|D_n|>\max(x^{1/q},z)]~qz^{q-1} \mathrm{d}z \\
&=& C \int_0^\infty \mathbf{P}[|D|>\max(x^{1/q},z)]~qz^{q-1} \mathrm{d}z \\
&\le& C\int_0^\infty \mathbf{P}[|D|>z]~qz^{q-1} \mathrm{d}z \\
&=& C\mathbf{E}[|D|^q\mathbf{1}_{[|D|^q>x]}].
\end{eqnarray}
By Lebesgue dominated convergence theorem, the right-hand side goes to $0$ as $x$ goes to infinity. Hence $(|D_n|^q)_{n \in \mathbf{N}}$ is uniformly integrable.
The third implication is classical. https://en.wikipedia.org/wiki/Uniform_integrability
Best Answer
The answer is negative. Suppose that for each $k>2$, the variable $X_k$ takes the values $\pm k^{1/p}$ with probability $k^{-1}$ each, and $P(X_k=0)=1-2k^{-1}$. (We can take $X_k=0$ for $k<2$.) Then $E[|X_k|^p]=2$ for all $k>2$. However, if $X$ satisfies $$\mathbb{P}(\vert X_k \vert > z) \le C \mathbb{P}(\vert X \vert > z)$$ for all $z > 0$ and $k\in \mathbb{N}$, then $$E[|X|^p]=\int_0^\infty\!\! P(|X|^p>x) \,dx \ge \sum_{k=3}^\infty \int_{k-1}^{k}\frac{P(|X_k|^p>x)}{C}=\frac2C\sum_{k=3}^\infty \frac1k=\infty \,.$$