This is certainly true if you choose $\lambda$ to be strictly smaller than the smaller eigenvalue of $DX(0)$. You may prove it inductively, by noticing that for a given $y$ the function $t\mapsto D^{\alpha}_y \Phi_t(y)$ solves a linear equation.
For instance, the first step goes as follows: the path of matrices $W(t):= D_y^{\alpha} \Phi_t(y)$ solves the ODE $$ W'(t) = DX(\Phi_t(y)) W(t), \quad W(0)=I, $$ where $\|DX(\Phi_t(y)) - DX(0)\| \leq C_0 e^{\lambda_0 t}$ for all $t\leq 0$. Then for every $\lambda_1<\lambda_0$ you can find $C_1$ such that $\|DX(\Phi_t(y))\| \leq C_1 e^{\lambda_1 t}$ for all $t\leq 0$.
A useful lemma for proving this and getting the uniformity you need is the following: given a continuous bounded path of matrices $t\mapsto A(t)$, $t\geq 0$, denote by $W_A(t)$ the solution of the linear Cauchy problem $$ W_A'(t) = A(t) W_A(t), \quad W_A(0) = I. $$ Assume that $\|X_A(t)X_A(s)^{-1}\|\leq \|W_A(t)W_A(s)^{-1}\|\leq c e^{\lambda (t-s)}$ for every $t\geq s\geq 0$. Then for every continuous bounded path of matrices $t\mapsto H(t)$, $t\geq 0$, there holds $$ \| X_{A+H}(t)X_{A+H}(s)^{-1}\|\leq W_{A+H}(t)W_{A+H}(s)^{-1}\|\leq c e^{\mu (t-s)}, \quad \forall t\geq s\geq 0, $$ with $\mu := \lambda + c \|H\|_{\infty}$.
(Sorry if here I switched to positive time, that's just because I am more used to work with stable manifolds).
