The answer is no: in general (and usually) there are no positive absolute constants $a$ and $b$ such that for some unit vector $v$ one has 
$$P(|X\cdot v|\ge a)\ge b.$$

Indeed, otherwise one would have $E(X\cdot v)^2\ge c:=ba^2>0$. However, if $X$ is uniformly distributed on the unit sphere in $\mathbb R^n$ and $v$ is a unit vector, then $(X\cdot v)^2$ has the beta distribution with parameters $1/2,(n-1)/2$ and hence $E(X\cdot v)^2=1/n<c$ if $n>1/c$. 

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The OP has changed the question, thus invalidating the above answer. Here is the answer to the changed question. 

Let $Y:=|X\cdot v|$, where $v$ is a unit vector corresponding  to the eigenvalue $s_1$. Then $0\le Y\le1$ and $EY^2=s_1$. So, for all $a\in(0,1)$ 
$$1(Y\ge a)\ge\frac{Y^2-a^2}{1-a^2}$$
and hence 
$$P(|X\cdot v|\ge a)=P(Y\ge a)\ge\frac{\max(0,s_1-a^2)}{1-a^2}.$$

This lower bound on $P(|X\cdot v|\ge a)$ is exact: It is attained if (i) $a^2\le s_1\le1$ and $(X\cdot v)^2$ only takes values $a^2$ and $1$ (with mean $E(X\cdot v)^2=s_1\in[a^2,1]$) or if (ii) $0\le s_1<a^2$ and $(X\cdot v)^2$ only takes value $s_1$.