The inequality is equivalent to $(1+t)^q\leqslant C\cdot(1+t^q)$, where $q=\frac12(p-2)$ hence $q\gt-\frac12$, and $t=x^2/\epsilon^2$ hence $t\gt0$. This cannot hold simultaneously for every $(t,p)$ since, for every fixed $t\gt0$, $(1+t)^q/(1+t^q)\to+\infty$ when $q\to+\infty$.
For every fixed $p$, let $C(p)$ denote the optimal constant such that this holds for every positive $t$. Then $C(p)=\left\{\begin{array}{cl}1 & \text{if}\ 2\lt p\leqslant4,\\ 2^{(p-4)/2}& \text{otherwise.} \end{array}\right. $ Note the discontinuity at $p=2$, from $C(2)=C(2^-)=\frac12$ to $C(2^+)=1$.
To prove this, consider the function $u_q:t\mapsto q\log(1+t)-\log(1+t^q)$ defined on $t\geqslant0$, then $u'_q(t)$ has the sign of $q\cdot(1-t^{q-1})$ hence $u_q(t)$ is maximal at $t=1$ if $q\gt1$ or $q\lt0$ and $u_q(t)$ is maximal at $t=0$ or $t=+\infty$ if $0\lt q\lt1$. The maximum is $(q-1)\log2=\frac12(p-4)\log2$ in the first case and $0$ in the second case. The values of $C(p)$ follow.
Edit: The OP asks in a comment whether, for some $p\gt1$ and $\epsilon\gt0$ fixed, there exists a finite constant $C_{p,\epsilon}$ such that $(\epsilon^2+x^2)^{(p-2)/2}\leqslant C_{\epsilon,p}+x^{p-2}$ for every positive $x$. The same approach as above yields the answer.
By homogeneity, $C$ is appropriate if and only if $C=D\,\epsilon^{p-2}$ where $(1+t)^q\leqslant D+t^q$ for every $t\geqslant0$, once again with $q=\frac12(p-2)$. Considering the function $v_q:t\mapsto(1+t)^q-t^q$ and its derivative, one sees that the supremum is $v_q(+\infty)=+\infty$ when $q\gt1$, $v_q(0)=1$ when $0\lt q\leqslant1$, $v_0(t)=0$ for any $t\gt0$ when $q=0$, and $v_q(+\infty)=1$ when $q\lt0$.
Thus, $C_{p,\epsilon}$ does not exist when $p\gt4$, $C_{p,\epsilon}=\epsilon^{p-2}$ when $p\leqslant4$ and $p\ne2$, and $C_{2,\epsilon}=0$. Finally, the optimal inequality one can achieve is the well-known fact that, for every $p\leqslant4$, $ (\epsilon^2+x^2)^{(p-2)/2}\leqslant \epsilon^{p-2}+x^{p-2}. $